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IGNOU BPHET-141 Solved Assignment | 1st January, 2025 to 31st December, 2025 | ELEMENTS OF MODERN PHYSICS | BSc CBCS Sample Solution

BPHET-141 Solved Assignment

PART A

a) A spacecraft $A$ has a speed 0.80 c with respect to the Earth. If the speed of another spacecraft $B$ with respect to spacecraft $A$ is 0.50 c , what is the speed of $B$ with respect to the Earth
(5)
b) A particle is travelling through the Earth’s atmosphere at a speed of 0.6 c . To an Earth-bound observer, the distance it travels is 4.0 km . How far does the particle travel in its own frame of reference?
c) A particle of rest mass 2.0 kg has an initial speed of $2 \times 10^{8} {ms}^{- 1}$ . A constant relativistic force of magnitude $1.5 \times 10^{6} N$ is exerted on the particle in the same direction as the initial relativistic momentum for 1000 s . Calculate the magnitudes of the initial and final relativistic linear momentum and its final speed.
d) A star emits light with wavelength 420 nm . An observer on earth measures the wavelength of the light received from the star to be 600 nm . Calculate the speed with which the star is moving.
a) Determine the wavelengths of the photons scattered at (i) $60^{\circ}$ and (ii) $90^{\circ}$ when $X$ rays of wavelength 4.5 pm are scattered from a target.
b) Calculate the de Broglie wavelength of $a^{87} R b$ atom that has been laser cooled to $200 μ K$ . (Assume that the kinetic energy is $\frac{3}{2} k_{B} T$ ).
c) Using Heisenberg’s Uncertainty Principle explain whether a particle trapped inside a one dimensional box of finite length can be at rest.
d) The quantum mechanical wave function for a particle is given by

ψ (x) = {\begin{array}{cc} N x^{\frac{3}{2}} e^{- 2 x}, & x > 0 \\ 0, & x < 0 \end{array}

Determine (i) the normalization constant

N

and(ii) the most probable position of the particle.
(10)

PART B
3. a) The eigenfunction of a particle confined in a box of length

L (0 \leq x \leq L)

ψ (x) = \sqrt{\frac{2}{L}} \sin (\frac{3 π x}{L})

Calculate

⟨ {\hat{p}}_{x}^{2} ⟩

and the probability of finding the particle between

x = 0

and

x = L / 4

b) A particle encounters a step potential of height

V_{0}

. What is the reflection and transmission coefficient if

E = 1.5 V_{0}

? Show that

R + T = 1

.
c) Show that, for a symmetric potential function

(V (x) = V (- x))

, the parity operator commutes with the Hamiltonian. What is the parity of the of the following eigenfunctions of a symmetric potential well of width 2a: (i)

ψ (x) = A \cos (\frac{3 π x}{2 a})

and (ii)

ψ (x) = A \sin (\frac{π x}{a})

a) The half life of $^{51} Cr$ is 27.70 days. After how many days will only $10 %$ of the element be left over?
(5)
b) Establish the relation for binding energy per nucleon for $_{Z}^{A}$ X nuclei. Calculate the value of binding energy per nucleon for $_{28}^{68} Ni$ . Given:

\begin{aligned} Mass of Ni : 63.9280 u \\ Mass of proton : 1.007825 u \\ Mass of neutron : 1.008665 u \end{aligned}

Is this nucleus stable?
(10)

c) A piece of wood from the ruins of an ancient dwelling was found to have a

^{14} C

activity of 14 disintegrations per minute per gram of its carbon content. The

^{14} C

activity of living wood is 16 disintegrations per minute per gram. How long ago did the tree die from which the wood sample came? Take half-life of radiocarbon as 5760 years.
d) Calculate

Q

-Value of the reaction:

_{13}^{27} Al +_{2}^{4} He \to_{14}^{30} Si +_{1}^{1} H

Take

m (_{13}^{27} Al) = 26.9815 u, m (_{1}^{1} H) = 1.0078 u, m (_{2}^{4} He) = 4.0026 u

and

_{14}^{30} Si = 29.9738 u

Answer:

Question:-1

a) A spacecraft $A$ has a speed 0.80 c with respect to the Earth. If the speed of another spacecraft $B$ with respect to spacecraft $A$ is 0.50 c, what is the speed of $B$ with respect to the Earth?

Answer:

We are tasked with finding the speed of spacecraft

B

with respect to the Earth when spacecraft

A

has a speed of 0.80c relative to the Earth and spacecraft

B

has a speed of 0.50c relative to spacecraft

A

To solve this problem, we will use the relativistic velocity addition formula, which accounts for the fact that velocities do not simply add up when the objects are moving at significant fractions of the speed of light.

Relativistic Velocity Addition Formula

The relativistic velocity addition formula for two velocities

u

and

v

in the same direction is given by:

u^{'} = \frac{u + v}{1 + \frac{u v}{c^{2}}}

where:

$u^{'}$ is the velocity of $B$ with respect to Earth,
$u$ is the velocity of $A$ with respect to Earth ( $0.80 c$ ),
$v$ is the velocity of $B$ with respect to $A$ ( $0.50 c$ ),
$c$ is the speed of light.

We need to use this formula to find the velocity of

B

with respect to the Earth.

Step 1: Apply the Formula

We substitute the given values into the formula:

u^{'} = \frac{0.80 c + 0.50 c}{1 + \frac{(0.80 c) (0.50 c)}{c^{2}}}

Step 2: Simplify the Expression

First, simplify the numerator and the denominator:

u^{'} = \frac{1.30 c}{1 + \frac{0.80 \times 0.50}{1}} = \frac{1.30 c}{1 + 0.40}

u^{'} = \frac{1.30 c}{1.40}

Step 3: Final Calculation

Now, divide

1.30

1.40

u^{'} = 0.92857 c

Thus, the speed of spacecraft

B

with respect to Earth is approximately:

u^{'} \approx 0.93 c

Conclusion

The speed of spacecraft

B

with respect to Earth is approximately

0.93 c

b) A particle is traveling through the Earth’s atmosphere at a speed of 0.6 c. To an Earth-bound observer, the distance it travels is 4.0 km. How far does the particle travel in its own frame of reference?

Answer:

To solve this problem, we need to apply length contraction, which is a relativistic effect that occurs when an object is moving at a significant fraction of the speed of light relative to an observer. The length contraction formula is:

L^{'} = L \sqrt{1 - \frac{v^{2}}{c^{2}}}

where:

$L^{'}$ is the length of the object (or distance traveled) in the object’s rest frame (i.e., the particle’s frame),
$L$ is the length of the object (or distance traveled) in the Earth-bound observer’s frame (4.0 km),
$v$ is the speed of the object relative to the observer (0.6c),
$c$ is the speed of light.

Step 1: Apply the Length Contraction Formula

Given:

$L = 4.0 km$ ,
$v = 0.6 c$ ,
$c = 3.0 \times 10^{8} m/s$ (speed of light, but it will cancel out in the equation).

We substitute these values into the length contraction formula:

L^{'} = 4.0 km \times \sqrt{1 - \frac{(0.6 c)^{2}}{c^{2}}}

Step 2: Simplify the Expression

First, simplify the term inside the square root:

L^{'} = 4.0 km \times \sqrt{1 - (0.6)^{2}}

L^{'} = 4.0 km \times \sqrt{1 - 0.36}

L^{'} = 4.0 km \times \sqrt{0.64}

L^{'} = 4.0 km \times 0.8

Step 3: Final Calculation

Now, multiply:

L^{'} = 3.2 km

Conclusion

In the particle’s own frame of reference, the distance it travels is 3.2 km. This is shorter than the distance observed by the Earth-bound observer due to the relativistic length contraction.

c) A particle of rest mass 2.0 kg has an initial speed of $2 \times 10^{8} {ms}^{- 1}$ . A constant relativistic force of magnitude $1.5 \times 10^{6} N$ is exerted on the particle in the same direction as the initial relativistic momentum for 1000 s. Calculate the magnitudes of the initial and final relativistic linear momentum and its final speed.

Answer:

Given Data:

Rest mass of the particle $m_{0} = 2.0 kg$
Initial speed $v_{i} = 2 \times 10^{8} m/s$
Relativistic force $F = 1.5 \times 10^{6} N$
Time $t = 1000 s$
Speed of light $c = 3.0 \times 10^{8} m/s$

We will calculate the initial relativistic linear momentum, final relativistic linear momentum, and the final speed of the particle.

Step 1: Calculate the Initial Relativistic Linear Momentum

The relativistic momentum

p

is given by the formula:

p = \frac{m_{0} v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

Substitute the given values:

Rest mass $m_{0} = 2.0 kg$ ,
Initial velocity $v_{i} = 2 \times 10^{8} m/s$ ,
Speed of light $c = 3.0 \times 10^{8} m/s$ .

We need to first calculate the Lorentz factor

γ_{i}

for the initial velocity

v_{i}

γ_{i} = \frac{1}{\sqrt{1 - \frac{v_{i}^{2}}{c^{2}}}}

Substitute

v_{i} = 2 \times 10^{8} m/s

γ_{i} = \frac{1}{\sqrt{1 - \frac{(2 \times 10^{8})^{2}}{(3 \times 10^{8})^{2}}}}

γ_{i} = \frac{1}{\sqrt{1 - \frac{4 \times 10^{16}}{9 \times 10^{16}}}} = \frac{1}{\sqrt{1 - 0.4444}} = \frac{1}{\sqrt{0.5556}} \approx 1.336

Now, calculate the initial relativistic momentum:

p_{i} = γ_{i} m_{0} v_{i} = 1.336 \times 2.0 kg \times 2 \times 10^{8} m/s

p_{i} = 1.336 \times 4.0 \times 10^{8} kg m/s

p_{i} = 5.344 \times 10^{8} kg m/s \approx 5.34 \times 10^{8} kg m/s

So, the initial relativistic momentum is:

p_{i} \approx 5.34 \times 10^{8} kg m/s

Step 2: Calculate the Change in Relativistic Momentum

We can find the change in relativistic momentum

Δ p

using the relation:

Δ p = F \cdot t

where:

$F = 1.5 \times 10^{6} N$ is the constant force,
$t = 1000 s$ is the duration of the force.

Substitute the given values:

Δ p = 1.5 \times 10^{6} N \times 1000 s = 1.5 \times 10^{9} kg m/s

Thus, the change in relativistic momentum is:

Δ p = 1.5 \times 10^{9} kg m/s

Step 3: Calculate the Final Relativistic Momentum

The final relativistic momentum

p_{f}

is the sum of the initial momentum

p_{i}

and the change in momentum

Δ p

p_{f} = p_{i} + Δ p

Substitute the values:

p_{f} = 5.34 \times 10^{8} kg m/s + 1.5 \times 10^{9} kg m/s

p_{f} = 2.03 \times 10^{9} kg m/s

Thus, the final relativistic momentum is:

p_{f} = 2.03 \times 10^{9} kg m/s

Step 4: Calculate the Final Speed of the Particle

To find the final speed

v_{f}

, we use the relativistic momentum formula:

p_{f} = \frac{m_{0} v_{f}}{\sqrt{1 - \frac{v_{f}^{2}}{c^{2}}}}

We can solve for

v_{f}

by rearranging this equation:

p_{f}^{2} = \frac{(m_{0} v_{f})^{2}}{1 - \frac{v_{f}^{2}}{c^{2}}}

Multiplying both sides by

1 - \frac{v_{f}^{2}}{c^{2}}

p_{f}^{2} (1 - \frac{v_{f}^{2}}{c^{2}}) = m_{0}^{2} v_{f}^{2}

Simplifying:

p_{f}^{2} - \frac{p_{f}^{2} v_{f}^{2}}{c^{2}} = m_{0}^{2} v_{f}^{2}

p_{f}^{2} = v_{f}^{2} (m_{0}^{2} + \frac{p_{f}^{2}}{c^{2}})

Solving for

v_{f}

v_{f} = \frac{p_{f}}{\sqrt{m_{0}^{2} + \frac{p_{f}^{2}}{c^{2}}}}

Substitute the known values:

v_{f} = \frac{2.03 \times 10^{9} kg m/s}{\sqrt{(2.0 kg)^{2} + \frac{(2.03 \times 10^{9} kg m/s)^{2}}{(3.0 \times 10^{8} m/s)^{2}}}}

First, calculate

\frac{p_{f}^{2}}{c^{2}}

\frac{(2.03 \times 10^{9})^{2}}{(3.0 \times 10^{8})^{2}} = \frac{4.1209 \times 10^{18}}{9 \times 10^{16}} = 45.79

Now, calculate the denominator:

m_{0}^{2} + \frac{p_{f}^{2}}{c^{2}} = (2.0)^{2} + 45.79 = 4.0 + 45.79 = 49.79

Thus, the final speed

v_{f}

is:

v_{f} = \frac{2.03 \times 10^{9}}{\sqrt{49.79}} = \frac{2.03 \times 10^{9}}{7.06} \approx 2.87 \times 10^{8} m/s

Thus, the final speed of the particle is:

v_{f} \approx 2.87 \times 10^{8} m/s

Summary of Results:

Initial relativistic momentum: $p_{i} \approx 5.34 \times 10^{8} kg m/s$
Final relativistic momentum: $p_{f} \approx 2.03 \times 10^{9} kg m/s$
Final speed: $v_{f} \approx 2.87 \times 10^{8} m/s$

These results show the changes in momentum and speed due to the application of a constant relativistic force over a time interval of 1000 seconds.

d) A star emits light with a wavelength of 420 nm. An observer on Earth measures the wavelength of the light received from the star to be 600 nm. Calculate the speed with which the star is moving.

Answer:

To solve this problem, we need to use the Doppler effect for light. The Doppler shift in the wavelength of light observed from a star depends on whether the star is moving toward or away from the observer. In this case, since the wavelength observed on Earth is longer than the emitted wavelength, it indicates that the star is moving away from the observer. This is known as redshift.

Given:

Emitted wavelength $λ_{emitted} = 420 nm$
Observed wavelength $λ_{observed} = 600 nm$
Speed of light $c = 3.0 \times 10^{8} m/s$

Step 1: Use the Doppler Formula for Relativistic Speeds

The relativistic Doppler shift formula is:

\frac{λ_{observed}}{λ_{emitted}} = \sqrt{\frac{1 + β}{1 - β}}

where:

$β = \frac{v}{c}$ is the ratio of the star’s velocity $v$ to the speed of light $c$ ,
$λ_{observed}$ is the observed wavelength,
$λ_{emitted}$ is the emitted wavelength.

We are given the wavelengths, and we need to solve for

v

, the velocity of the star.

Step 2: Rearrange the Doppler Formula

Rearranging the formula to solve for

β

\frac{λ_{observed}}{λ_{emitted}} = \sqrt{\frac{1 + β}{1 - β}}

Squaring both sides:

{(\frac{λ_{observed}}{λ_{emitted}})}^{2} = \frac{1 + β}{1 - β}

\frac{1 + β}{1 - β} = {(\frac{λ_{observed}}{λ_{emitted}})}^{2}

Substitute the values:

\frac{1 + β}{1 - β} = {(\frac{600}{420})}^{2} = {(1.4286)}^{2} = 2.0408

Step 3: Solve for $β$

Now we can solve for

β

1 + β = 2.0408 (1 - β)

Expanding the right-hand side:

1 + β = 2.0408 - 2.0408 β

Move the terms involving

β

to one side:

1 + β + 2.0408 β = 2.0408

1 + 3.0408 β = 2.0408

Now subtract 1 from both sides:

3.0408 β = 1.0408

Solve for

β

β = \frac{1.0408}{3.0408} \approx 0.342

Step 4: Calculate the Speed of the Star

Now, we can calculate the velocity

v

of the star using

β = \frac{v}{c}

β = \frac{v}{c}

v = β \times c = 0.342 \times 3.0 \times 10^{8} m/s

v \approx 1.026 \times 10^{8} m/s

Final Answer:

The speed of the star is approximately:

v \approx 1.03 \times 10^{8} m/s

Back to Course

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IGNOU BPHET-141 Solved Assignment | 1st January, 2025 to 31st December, 2025 | ELEMENTS OF MODERN PHYSICS | BSc CBCS

Sample Solution

BPHET-141 Solved Assignment

Answer:

Question:-1

a) A spacecraft A A AAA has a speed 0.80 c with respect to the Earth. If the speed of another spacecraft B B BBB with respect to spacecraft A A AAA is 0.50 c, what is the speed of B B BBB with respect to the Earth?

Answer:

Relativistic Velocity Addition Formula

Step 1: Apply the Formula

Step 2: Simplify the Expression

Step 3: Final Calculation

Conclusion

b) A particle is traveling through the Earth’s atmosphere at a speed of 0.6 c. To an Earth-bound observer, the distance it travels is 4.0 km. How far does the particle travel in its own frame of reference?

Answer:

Step 1: Apply the Length Contraction Formula

Step 2: Simplify the Expression

Step 3: Final Calculation

Conclusion

Answer:

Given Data:

Step 1: Calculate the Initial Relativistic Linear Momentum

Step 2: Calculate the Change in Relativistic Momentum

Step 3: Calculate the Final Relativistic Momentum

Step 4: Calculate the Final Speed of the Particle

Summary of Results:

d) A star emits light with a wavelength of 420 nm. An observer on Earth measures the wavelength of the light received from the star to be 600 nm. Calculate the speed with which the star is moving.

Answer:

Given:

Step 1: Use the Doppler Formula for Relativistic Speeds

Step 2: Rearrange the Doppler Formula

Step 3: Solve for β β beta\betaβ

Step 4: Calculate the Speed of the Star

Final Answer:

a) A spacecraft $A$ has a speed 0.80 c with respect to the Earth. If the speed of another spacecraft $B$ with respect to spacecraft $A$ is 0.50 c, what is the speed of $B$ with respect to the Earth?

Step 3: Solve for $β$