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IGNOU BPHET-141 Solved Assignment 2024 | B.Sc (G) CBCS

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IGNOU BPHET-141 Assignment Question Paper 2024

bphet-141-solved-assignment-2024–qp-ce9527ef-ea07-4607-a728-90a5edc19fd6

bphet-141-solved-assignment-2024–qp-ce9527ef-ea07-4607-a728-90a5edc19fd6

BPHET-141 Solved Assignment 2024 QP
PART A
  1. a) Light takes 5.2 years to travel from a distant planet to earth. If the astronaut travelled from earth to that planet at a speed of 0.8 c 0.8 c 0.8c0.8 \mathrm{c}0.8c, how long would it take according to the astronaut’s clock, to reach the planet?
    b) Calculate the linear momentum, total energy (in M e V M e V MeV\mathrm{MeV}MeV ) and kinetic energy (in M e V M e V MeV\mathrm{MeV}MeV ) of a particle travelling at 0.5 c 0.5 c 0.5c0.5 \mathrm{c}0.5c, given that its rest mass is 938 M e V 938 M e V 938MeV938 \mathrm{MeV}938MeV.
    c) Two spaceships of proper length L 0 L 0 L_(0)L_0L0 approach the earth from opposite direction at velocities ± 0.7 c ± 0.7 c +-0.7 c\pm 0.7 c±0.7c. What is the length of one of the spaceships with respect to the other?
    d) Calculate the relativistic mass and the rest mass of a negatively charged particle which has linear momentum of magnitude 1.80 × 10 − 21 k g m s − 1 1.80 × 10 − 21 k g m s − 1 1.80 xx10^(-21)kgms^(-1)1.80 \times 10^{-21} \mathrm{kgms}^{-1}1.80×10−21kgms−1 at a speed of 0.9 c 0.9 c 0.9c0.9 \mathrm{c}0.9c.
    e) Derive the relativistic energy – momentum relation for a free particle.
  2. a) Light of frequency 6.0 × 10 14 H z 6.0 × 10 14 H z 6.0 xx10^(14)Hz6.0 \times 10^{14} \mathrm{~Hz}6.0×1014 Hz is incident on a metal surface. Electrons with a maximum speed of 4.5 × 10 5 m s − 1 4.5 × 10 5 m s − 1 4.5 xx10^(5)ms^(-1)4.5 \times 10^5 \mathrm{~ms}^{-1}4.5×105 ms−1 are ejected from the surface. Calculate the work function and the cut-off frequency.
    b) Determine the wavelengths of a photon and an electron having energy 5.0 × 10 3 5.0 × 10 3 5.0 xx10^(3)5.0 \times 10^35.0×103 e V e V eV\mathrm{eV}eV and 5.5 × 10 3 e V 5.5 × 10 3 e V 5.5 xx10^(3)eV5.5 \times 10^3 \mathrm{eV}5.5×103eV respectively. Which of these would you use to probe atomic structures?
    c) Use the uncertainty principle to estimate the approximate size of an atom.
    d) The wave function of a particle of mass m m mmm that can move freely between − L ≤ x ≤ L − L ≤ x ≤ L -L <= x <= L-L \leq x \leq L−L≤x≤L is given as:
ψ ( x , t ) = { A cos ⁡ ( 2 π x L ) exp ⁡ ( − i E t ℏ ) for − L < x < L 0 elsewhere ψ ( x , t ) = A cos ⁡ 2 π x L exp ⁡ − i E t ℏ for − L < x < L 0 elsewhere psi(x,t)={[A cos((2pi x)/(L))exp(-(iEt)/(ℏ)),” for “-L < x < L],[0,” elsewhere “]:}\psi(x, t)=\left\{\begin{array}{cc} A \cos \left(\frac{2 \pi x}{L}\right) \exp \left(-\frac{i E t}{\hbar}\right) & \text { for }-L<x<L \\ 0 & \text { elsewhere } \end{array}\right.ψ(x,t)={Acos⁡(2πxL)exp⁡(−iEtℏ) for −L<x<L0 elsewhere
Using time dependent Schrodinger equation, determine the value of the constant E.
e) Show that [ L ^ y , L ^ x ] = − i ℏ L ^ z L ^ y , L ^ x = − i ℏ L ^ z [ hat(L)_(y), hat(L)_(x)]=-iℏ hat(L)_(z)\left[\hat{L}_y, \hat{L}_x\right]=-i \hbar \hat{L}_z[L^y,L^x]=−iℏL^z.
PART B
3. a) The eigenfunction of a particle confined in a box of length L ( 0 ≤ x ≤ L ) L ( 0 ≤ x ≤ L ) L(0 <= x <= L)L(0 \leq x \leq L)L(0≤x≤L) is
ψ ( x ) = A sin ⁡ ( 4 π x L ) ψ ( x ) = A sin ⁡ 4 π x L psi(x)=A sin((4pi x)/(L))\psi(x)=A \sin \left(\frac{4 \pi x}{L}\right)ψ(x)=Asin⁡(4πxL)
Determine the normalization constant A A AAA, the expectation value of the kinetic energy and the probability of finding the particle between x = L / 2 x = L / 2 x=L//2x=L / 2x=L/2 and x = 3 L / 4 x = 3 L / 4 x=3L//4x=3 L / 4x=3L/4.
b) A particle of energy 4 e V 4 e V 4eV4 \mathrm{eV}4eV is incident of a potential with energy 8 e V 8 e V 8eV8 \mathrm{eV}8eV. Calculate the distance in which the probability of finding the particle becomes 0.04 as it goes into the "forbidden" region.
c) Calculate the penetration depth for an electron of energy 30 e V 30 e V 30eV30 \mathrm{eV}30eV, trapped by an electrostatic potential 80 e V 80 e V 80eV80 \mathrm{eV}80eV. How much energy would be required to "free" the electron from the well?
4. a) Plot a graph of the binding energy per nucleon as a function of mass number. Write the main features of the curve.
b) The mean life of an element is 18 months. Determine the time required for 60 % 60 % 60%60 \%60% of the element to decay.
c) Calculate the kinetic energy of an alpha particle emitted in the alpha decay of uranium isotope 92 233 U → 90 229 T h + α 92 233 U → 90 229 T h + α _(92)^(233)U rarr_(90)^(229)Th+alpha{ }_{92}^{233} U \rightarrow{ }_{90}^{229} T h+\alpha92233U→90229Th+α particle . It is given that
m ( 92 233 U ) = 233.0396343 u , m ( 90 229 T h ) = 229.03176 u and m ( 2 4 H ) = 4.002603 u . m 92 233 U = 233.0396343 u , m 90 229 T h = 229.03176 u and m 2 4 H = 4.002603 u . m(_(92)^(233)U)=233.0396343u,m(_(90)^(229)Th)=229.03176u” and “m(_(2)^(4)H)=4.002603u”. “m\left({ }_{92}^{233} U\right)=233.0396343 \mathrm{u}, m\left({ }_{90}^{229} T h\right)=229.03176 \mathrm{u} \text { and } m\left({ }_2^4 H\right)=4.002603 \mathrm{u} \text {. }m(92233U)=233.0396343u,m(90229Th)=229.03176u and m(24H)=4.002603u.
d) Calculate the energy of reaction ( Q − V a l u e Q − V a l u e Q-Value\mathrm{Q}-\mathrm{Value}Q−Value ) in M e V M e V MeV\mathrm{MeV}MeV in the fusion reaction of Deuterium-Tritium into helium:
1 2 H + 1 3 H → 2 4 H e + n 1 2 H + 1 3 H → 2 4 H e + n _(1)^(2)H+_(1)^(3)Hrarr_(2)^(4)He+n{ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{n}12H+13H→24He+n
Take m ( 1 2 H ) = 2.0141029 u , m ( 1 3 H ) = 3.016049 u , m ( 2 4 H e ) = 4.0026 u m 1 2 H = 2.0141029 u , m 1 3 H = 3.016049 u , m 2 4 H e = 4.0026 u m(_(1)^(2)H)=2.0141029u,m(_(1)^(3)H)=3.016049u,m(_(2)^(4)He)=4.0026um\left({ }_1^2 H\right)=2.0141029 \mathrm{u}, m\left({ }_1^3 H\right)=3.016049 \mathrm{u}, m\left({ }_2^4 H e\right)=4.0026 \mathrm{u}m(12H)=2.0141029u,m(13H)=3.016049u,m(24He)=4.0026u and m ( n ) = 1.008665 u m ( n ) = 1.008665 u m(n)=1.008665um(n)=1.008665 \mathrm{u}m(n)=1.008665u.
e) The half-life of 232 T h 232 T h ^(232)Th{ }^{232} \mathrm{Th}232Th is known to be 14 × 10 9 y r s 14 × 10 9 y r s 14 xx10^(9)yrs14 \times 10^9 \mathrm{yrs}14×109yrs. Calculate the decay constant (in s − 1 s − 1 s^(-1)\mathrm{s}^{-1}s−1 ) and the rate of disintegration for 1 g 1 g 1g1 \mathrm{~g}1 g of Thorium.
\(cos\left(2\theta \right)=cos^2\theta -sin^2\theta \)

BPHET-141 Sample Solution 2024

bphet-141-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

bphet-141-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

BPHET-141 Solved Assignment 2024 SS
PART A
  1. a) Light takes 5.2 years to travel from a distant planet to earth. If the astronaut travelled from earth to that planet at a speed of 0.8 c 0.8 c 0.8c0.8 \mathrm{c}0.8c, how long would it take according to the astronaut’s clock, to reach the planet?
Answer:
To find the time it takes for the astronaut to reach the planet according to the astronaut’s clock, we need to consider time dilation due to special relativity. The time measured by the astronaut, t ′ t ′ t^(‘)t’t′, is related to the time measured by an observer on Earth, t t ttt, by the following equation:
t ′ = t γ t ′ = t γ t^(‘)=(t)/( gamma)t’ = \frac{t}{\gamma}t′=tγ
where γ γ gamma\gammaγ is the Lorentz factor, given by:
γ = 1 1 − ( v c ) 2 γ = 1 1 − v c 2 gamma=(1)/(sqrt(1-((v)/(c))^(2)))\gamma = \frac{1}{\sqrt{1 – \left(\frac{v}{c}\right)^2}}γ=11−(vc)2
Here, v = 0.8 c v = 0.8 c v=0.8 cv = 0.8cv=0.8c is the speed of the astronaut, and c c ccc is the speed of light. The time it takes for light to travel from the planet to Earth is given as 5.2 years, so t = 5.2 years t = 5.2 years t=5.2″ years”t = 5.2 \text{ years}t=5.2 years.
First, let’s calculate the Lorentz factor:
γ = 1 1 − ( 0.8 ) 2 = 1 1 − 0.64 = 1 0.36 = 1 0.6 ≈ 1.667 γ = 1 1 − ( 0.8 ) 2 = 1 1 − 0.64 = 1 0.36 = 1 0.6 ≈ 1.667 gamma=(1)/(sqrt(1-(0.8)^(2)))=(1)/(sqrt(1-0.64))=(1)/(sqrt0.36)=(1)/(0.6)~~1.667\gamma = \frac{1}{\sqrt{1 – (0.8)^2}} = \frac{1}{\sqrt{1 – 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.667γ=11−(0.8)2=11−0.64=10.36=10.6≈1.667
Now, we can calculate the time it takes according to the astronaut’s clock:
t ′ = t γ = 5.2 years 1.667 ≈ 3.12 years t ′ = t γ = 5.2 years 1.667 ≈ 3.12 years t^(‘)=(t)/( gamma)=(5.2″ years”)/(1.667)~~3.12″ years”t’ = \frac{t}{\gamma} = \frac{5.2 \text{ years}}{1.667} \approx 3.12 \text{ years}t′=tγ=5.2 years1.667≈3.12 years
So, according to the astronaut’s clock, it would take approximately 3.12 years to reach the planet.
b) Calculate the linear momentum, total energy (in M e V M e V MeV\mathrm{MeV}MeV ) and kinetic energy (in M e V M e V MeV\mathrm{MeV}MeV ) of a particle travelling at 0.5 c 0.5 c 0.5c0.5 \mathrm{c}0.5c, given that its rest mass is 938 M e V 938 M e V 938MeV938 \mathrm{MeV}938MeV.
Answer:
To calculate the linear momentum, total energy, and kinetic energy of a particle traveling at 0.5 c 0.5 c 0.5 c0.5c0.5c, where c c ccc is the speed of light, and given that its rest mass is 938 M e V 938 M e V 938MeV938 \mathrm{MeV}938MeV (which corresponds to the rest mass of a proton), we can use the following relativistic formulas:
  1. Linear Momentum ( p p ppp):
    p = γ m 0 v p = γ m 0 v p=gammam_(0)vp = \gamma m_0 vp=γm0v
    where m 0 m 0 m_(0)m_0m0 is the rest mass, v v vvv is the velocity of the particle, and γ γ gamma\gammaγ is the Lorentz factor, given by:
    γ = 1 1 − v 2 c 2 γ = 1 1 − v 2 c 2 gamma=(1)/(sqrt(1-(v^(2))/(c^(2))))\gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}γ=11−v2c2
  2. Total Energy ( E E EEE):
    E = γ m 0 c 2 E = γ m 0 c 2 E=gammam_(0)c^(2)E = \gamma m_0 c^2E=γm0c2
  3. Kinetic Energy ( K K KKK):
    K = E − m 0 c 2 = ( γ − 1 ) m 0 c 2 K = E − m 0 c 2 = ( γ − 1 ) m 0 c 2 K=E-m_(0)c^(2)=(gamma-1)m_(0)c^(2)K = E – m_0 c^2 = (\gamma – 1) m_0 c^2K=E−m0c2=(γ−1)m0c2
Let’s calculate each of these quantities:
  1. Linear Momentum:
    γ = 1 1 − ( 0.5 c ) 2 c 2 = 1 1 − 0.25 = 1 0.75 = 1 0.866 ≈ 1.155 γ = 1 1 − ( 0.5 c ) 2 c 2 = 1 1 − 0.25 = 1 0.75 = 1 0.866 ≈ 1.155 gamma=(1)/(sqrt(1-((0.5 c)^(2))/(c^(2))))=(1)/(sqrt(1-0.25))=(1)/(sqrt0.75)=(1)/(0.866)~~1.155\gamma = \frac{1}{\sqrt{1 – \frac{(0.5c)^2}{c^2}}} = \frac{1}{\sqrt{1 – 0.25}} = \frac{1}{\sqrt{0.75}} = \frac{1}{0.866} \approx 1.155γ=11−(0.5c)2c2=11−0.25=10.75=10.866≈1.155
    p = ( 1.155 ) ( 938 M e V ) ( 0.5 ) = 542.09 M e V / c p = ( 1.155 ) ( 938 M e V ) ( 0.5 ) = 542.09 M e V / c p=(1.155)(938MeV)(0.5)=542.09MeV//cp = (1.155)(938 \mathrm{MeV})(0.5) = 542.09 \mathrm{MeV/c}p=(1.155)(938MeV)(0.5)=542.09MeV/c
  2. Total Energy:
    E = ( 1.155 ) ( 938 M e V ) = 1082.99 M e V E = ( 1.155 ) ( 938 M e V ) = 1082.99 M e V E=(1.155)(938MeV)=1082.99MeVE = (1.155)(938 \mathrm{MeV}) = 1082.99 \mathrm{MeV}E=(1.155)(938MeV)=1082.99MeV
  3. Kinetic Energy:
    K = ( 1.155 − 1 ) ( 938 M e V ) = 0.155 × 938 M e V = 145.39 M e V K = ( 1.155 − 1 ) ( 938 M e V ) = 0.155 × 938 M e V = 145.39 M e V K=(1.155-1)(938MeV)=0.155 xx938MeV=145.39MeVK = (1.155 – 1)(938 \mathrm{MeV}) = 0.155 \times 938 \mathrm{MeV} = 145.39 \mathrm{MeV}K=(1.155−1)(938MeV)=0.155×938MeV=145.39MeV
Therefore, the linear momentum of the particle is approximately 542.09 M e V / c 542.09 M e V / c 542.09MeV//c542.09 \mathrm{MeV/c}542.09MeV/c, the total energy is approximately 1082.99 M e V 1082.99 M e V 1082.99MeV1082.99 \mathrm{MeV}1082.99MeV, and the kinetic energy is approximately 145.39 M e V 145.39 M e V 145.39MeV145.39 \mathrm{MeV}145.39MeV.
c) Two spaceships of proper length L 0 L 0 L_(0)L_0L0 approach the earth from opposite direction at velocities ± 0.7 c ± 0.7 c +-0.7 c\pm 0.7 c±0.7c. What is the length of one of the spaceships with respect to the other?
Answer:
To find the length of one of the spaceships with respect to the other, we need to consider the relativistic effect of length contraction. When an object is moving relative to an observer, its length appears contracted along the direction of motion according to the formula:
L = L 0 1 − v 2 c 2 L = L 0 1 − v 2 c 2 L=L_(0)sqrt(1-(v^(2))/(c^(2)))L = L_0 \sqrt{1 – \frac{v^2}{c^2}}L=L01−v2c2
where L L LLL is the observed length, L 0 L 0 L_(0)L_0L0 is the proper length (the length of the object in its rest frame), v v vvv is the relative velocity between the observer and the object, and c c ccc is the speed of light.
In this case, the relative velocity between the two spaceships is the sum of their individual velocities, as they are moving in opposite directions. Therefore, v = 0.7 c + 0.7 c = 1.4 c v = 0.7 c + 0.7 c = 1.4 c v=0.7 c+0.7 c=1.4 cv = 0.7c + 0.7c = 1.4cv=0.7c+0.7c=1.4c. However, the maximum possible relative velocity cannot exceed the speed of light, so we need to use the relativistic velocity addition formula to find the correct relative velocity:
v rel = v 1 + v 2 1 + v 1 v 2 c 2 = 0.7 c + 0.7 c 1 + ( 0.7 c ) ( 0.7 c ) c 2 = 1.4 c 1 + 0.49 = 1.4 c 1.49 ≈ 0.9396 c v rel = v 1 + v 2 1 + v 1 v 2 c 2 = 0.7 c + 0.7 c 1 + ( 0.7 c ) ( 0.7 c ) c 2 = 1.4 c 1 + 0.49 = 1.4 c 1.49 ≈ 0.9396 c v_(“rel”)=(v_(1)+v_(2))/(1+(v_(1)v_(2))/(c^(2)))=(0.7 c+0.7 c)/(1+((0.7 c)(0.7 c))/(c^(2)))=(1.4 c)/(1+0.49)=(1.4 c)/(1.49)~~0.9396 cv_{\text{rel}} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} = \frac{0.7c + 0.7c}{1 + \frac{(0.7c)(0.7c)}{c^2}} = \frac{1.4c}{1 + 0.49} = \frac{1.4c}{1.49} \approx 0.9396cvrel=v1+v21+v1v2c2=0.7c+0.7c1+(0.7c)(0.7c)c2=1.4c1+0.49=1.4c1.49≈0.9396c
Now, we can use the length contraction formula to find the observed length of one spaceship with respect to the other:
L = L 0 1 − ( 0.9396 c ) 2 c 2 = L 0 1 − 0.8829 = L 0 0.1171 ≈ 0.3421 L 0 L = L 0 1 − ( 0.9396 c ) 2 c 2 = L 0 1 − 0.8829 = L 0 0.1171 ≈ 0.3421 L 0 L=L_(0)sqrt(1-((0.9396 c)^(2))/(c^(2)))=L_(0)sqrt(1-0.8829)=L_(0)sqrt0.1171~~0.3421L_(0)L = L_0 \sqrt{1 – \frac{(0.9396c)^2}{c^2}} = L_0 \sqrt{1 – 0.8829} = L_0 \sqrt{0.1171} \approx 0.3421 L_0L=L01−(0.9396c)2c2=L01−0.8829=L00.1171≈0.3421L0
So, the length of one of the spaceships with respect to the other is approximately 0.3421 L 0 0.3421 L 0 0.3421L_(0)0.3421 L_00.3421L0, meaning it appears to be about 34.21% of its proper length.
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