Modern Physics Numerical Solutions

phy-1109 PHY-1109 Physics

Topic: Modern Physics (Relativity, Quantum Mechanics, Nuclear Physics) Source: PYQs 2016–2024

II. MODERN PHYSICS

	Subtopic	Question (Numerical Input/Calculation Required)	Year(s)
1	Relativistic Mass	A man has a mass of 100 Kg on the ground. When he is in a rocket ship in flight, his mass is 101 Kg… What is the speed of the rocket ship?.	2017
2	Relativistic Mass	Man 70 Kg (ground). 71 Kg (in rocket). What is speed of rocket ship?.	2020
3	Length Contraction	A rocket ship is 100 m long on the ground. When it is in flight, its length is 99 m… What is its speed?.	2018, 2019
4	Photoelectric Effect	The photoelectric threshold frequency of silver is $0.86\times 10^{15}$ Hz. Calculate (i) the maximum kinetic energy, (ii) the stopping potential for $4\times 10^{15}$ Hz UV light.	2019
5	Photoelectric Effect	Metal threshold wavelength $6000\text{A˚}$. Calculate stopping potential when irradiated with (i) monochromatic light $\lambda =4000\text{A˚}$, (ii) light having twice the intensity and frequency.	2024
6	Compton Scattering	X-ray photon of wavelength 0.3 $\text{A˚}$ undergoes a 60° Compton scattering. Find the wavelength of scattered photon and kinetic energy imparted to the recoiling electron.	2016
7	Compton Scattering	X-ray scattered by free electron. At 45°, ${\lambda }^{\prime }=0.22\text{A˚}$. What is the wavelength of direct beam?.	2017
8	De-Broglie Waves	An electron has de-Broglie wavelength of 2 PM. Find its Kinetic energy and the phase & group velocities of its de-Broglie waves.	2021
9	De-Broglie Waves	Calculate the wavelength associated with a thermal neutron of energy 0.025 eV.	2022
10	Bohr Model (Wavelength)	Find the wavelength of the spectral line corresponding to the transition in the hydrogen atom from $n=4$ to $n=2$ state.	2020
11	Bohr Model (Energy)	How much energy is required to remove an electron in the $n=2$ state from a Hydrogen atom?.	2018
12	Radioactive Decay	Half-life of ${\text{U}}^{238}$ is $4.51\times 10^9$ years. What percentage of ${\text{U}}^{238}$ that existed $10^{10}$ years ago still survives?.	2019, 2021, 2022
13	Radioactive Decay	Half life of Radium is 1600 years. In how many years will $1$ gm of pure radium lose (i) one centigram, and (ii) be reduced to $1$ centigram.	2020, 2024
14	Radioactive Activity	Half-life of Radium is 26.8 minutes. Calculate the amount of ${\text{Ra}}^{214}$ which activity is 1 curie.	2017
15	Radioactive Decay	Half-life of Radon is 3.82 day. Find the time of decay to 60% of it.	2018
16	Nuclear Energy	Calculate the mass defect and binding energy of a deuteron.	2016
17	Nuclear Energy	Calculate the binding energy of a neutron in the ${}_3^7\text{Li}$ nucleus. Express the result in joules.	2023
18	Nuclear Fission	Calculate the energy released in a fission of ${}^{235}\text{U}$ nucleus.	2016
19	Nuclear Power	Calculate rate of consumption of ${\text{U}}^{235}$ per year for Roppur Nuclear Power Plant operating at 250 megawatts.	2024
20	Uncertainty Principle	Lifetime of an excited state $10^{-8}$ s. Calculate the minimum uncertainty in the energy of the excited state.	2023

1. Special Theory of Relativity

Q1: Relativistic Mass (2017)

Question: Mass is $100$ kg on ground ($m_0$), $101$ kg in flight ($m$). Find speed ($v$).

Formula: $m=\frac{m_0}{\sqrt{1-v^2/c^2}}$

Calculation:

101 = \frac{100}{\sqrt{1 - v^2/c^2}} \Rightarrow \sqrt{1 - \frac{v^2}{c^2}} = \frac{100}{101}$$$$1 - \frac{v^2}{c^2} = \left(\frac{100}{101}\right)^2 \approx 0.9803$$$$\frac{v^2}{c^2} = 1 - 0.9803 = 0.0197$$$$v = c\sqrt{0.0197} \approx 0.14c

Answer: $4.2\times 10^7\text{ m/s}$ (or $0.14c$)

Q2: Relativistic Mass (2020)

Question: Mass $70$ kg ($m_0$) becomes $71$ kg ($m$). Find speed.

Calculation:

\sqrt{1 - \frac{v^2}{c^2}} = \frac{70}{71} \approx 0.9859$$$$\frac{v^2}{c^2} = 1 - (0.9859)^2 = 1 - 0.972 = 0.028$$$$v = \sqrt{0.028}c \approx 0.167c

Answer: $5\times 10^7\text{ m/s}$ (or $0.167c$)

Q3: Length Contraction (2018, 2019)

Question: $L_0=100$ m (ground), $L=99$ m (flight). Find speed.

Formula: $L=L_0\sqrt{1-v^2/c^2}$

Calculation:

99 = 100 \sqrt{1 - \frac{v^2}{c^2}} \Rightarrow 0.99 = \sqrt{1 - \frac{v^2}{c^2}}$$$$0.99^2 = 1 - \frac{v^2}{c^2} \Rightarrow 0.9801 = 1 - \frac{v^2}{c^2}$$$$\frac{v^2}{c^2} = 0.0199$$$$v = \sqrt{0.0199}c \approx 0.141c

Answer: $4.23\times 10^7\text{ m/s}$ (or $0.141c$)

2. Quantum Mechanics (Photoelectric & Compton)

Q4: Photoelectric Effect (2019)

Given: ${\nu }_0=0.86\times 10^{15}$ Hz, Light $\nu =4\times 10^{15}$ Hz.

Formula: $K_{\text{max}}=h(\nu -{\nu }_0)$ and $V_s=K_{\text{max}}/e$

Calculation:

(i) Max Kinetic Energy:

K_{\text{max}} = 6.63 \times 10^{-34} (4 - 0.86) \times 10^{15}$$$$K_{\text{max}} = 6.63 \times 3.14 \times 10^{-19} \approx 20.8 \times 10^{-19} \text{ J}

(ii) Stopping Potential:

$V_s=\frac{20.8\times 10^{-19}\text{ J}}{1.6\times 10^{-19}\text{ C}}\approx 13\text{ V}$

Answer: (i) $2.08\times 10^{-18}\text{ J}$, (ii) $13\text{ V}$

Q5: Photoelectric Effect (2024)

Given: ${\lambda }_0=6000\text{A˚}$. Incident $\lambda =4000\text{A˚}$.

Formula: $e{V}_s=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)$

Using $hc\approx 12400\text{ eV}\cdot \text{A˚}$:

Calculation:

(i) For $4000\text{A˚}$:

eV_s = 12400 \left(\frac{1}{4000} - \frac{1}{6000}\right) = 12400 \left(\frac{3-2}{12000}\right)$$$$eV_s = 12400 \times \frac{1}{12000} \approx 1.033 \text{ eV} \Rightarrow V_s = 1.033 \text{ V}

(ii) Light with twice intensity and twice frequency:

• Twice Intensity: Does not change Stopping Potential (only saturation current).

• Twice Frequency: New ${\nu }^{\prime }=2\nu \Rightarrow$ New ${\lambda }^{\prime }=\lambda /2=2000\text{A˚}$.

  $e{V}_s^{\prime }=12400\left(\frac{1}{2000}-\frac{1}{6000}\right)=12400\left(\frac{3-1}{6000}\right)=12400\left(\frac{2}{6000}\right)\approx 4.13\text{ eV}$

  Answer: (i) $1.03\text{ V}$; (ii) $4.13\text{ V}$ (Intensity change has no effect on $V_s$).

Q6: Compton Scattering (2016)

Given: $\lambda =0.3\text{A˚}$, $\theta =60^{\circ }$.

Formula: $\Delta \lambda =\frac{h}{m_{0}c}(1-\cos \theta )=0.0243\text{A˚}(1-\cos 60^{\circ })$

Calculation:

\Delta \lambda = 0.0243 (1 - 0.5) = 0.01215 \text{\AA}$$$$\lambda' = 0.3 + 0.01215 = 0.31215 \text{\AA}

Recoil Energy $K_e=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }})$:

$K_e\approx 12400\left(\frac{1}{0.3}-\frac{1}{0.31215}\right)=12400(3.333-3.203)\approx 1612\text{ eV}$

Answer: ${\lambda }^{\prime }=0.312\text{A˚}$, $K_e=1.61\text{ keV}$

Q7: Compton Scattering (2017)

Given: $\theta =45^{\circ }$, Scattered ${\lambda }^{\prime }=0.22\text{A˚}$.

Calculation:

$\Delta \lambda =0.0243(1-\cos 45^{\circ })=0.0243(1-0.707)\approx 0.0071\text{A˚}$

Direct beam $\lambda ={\lambda }^{\prime }-\Delta \lambda =0.22-0.0071=0.2129\text{A˚}$

Answer: $0.213\text{A˚}$

3. Matter Waves & Atomic Model

Q8: De-Broglie Waves (2021)

Given: Electron $\lambda =2\text{ pm}=0.002\text{ nm}$.

Analysis: Since $\lambda$ is extremely small (high energy), we must use relativistic formulas.

$pc=\frac{hc}{\lambda }=\frac{1240\text{ eV nm}}{0.002\text{ nm}}=620,000\text{ eV}=0.62\text{ MeV}$.

Rest mass energy $E_0=0.511\text{ MeV}$. Since $pc\approx E_0$, it is relativistic.

Calculation:

(i) Kinetic Energy: $E_{total}=\sqrt{(pc{)}^2+(m_0{c}^2{)}^2}=\sqrt{0.62^2+0.511^2}\approx 0.803\text{ MeV}$.

$K=E_{total}-E_0=0.803-0.511=0.292\text{ MeV}$.

(ii) Velocities:

Particle/Group Velocity $v_g=v$: Using $pc=E_{total}\frac{v}{c}\Rightarrow \frac{v}{c}=\frac{pc}{E_{total}}=\frac{0.62}{0.803}\approx 0.77c$.

Phase Velocity $v_p=\frac{c^2}{v}=\frac{c}{0.77}\approx 1.3c$.

Answer: $K=292\text{ keV}$, $v_g=0.77c$, $v_p=1.3c$

Q9: De-Broglie Waves (2022)

Given: Thermal neutron $K=0.025\text{ eV}$.

Formula: $\lambda =\frac{h}{\sqrt{2mK}}$ (Non-relativistic is fine here).

For neutrons, simplified formula: $\lambda (\text{A˚})\approx \frac{0.286}{\sqrt{K(\text{eV})}}$.

Calculation:

$\lambda =\frac{0.286}{\sqrt{0.025}}=\frac{0.286}{0.158}\approx 1.81\text{A˚}$

Answer: $1.81\text{A˚}$

Q10: Bohr Model (2020)

Question: Hydrogen $n=4\to n=2$ (H-$\beta$ line).

Formula: $\frac{1}{\lambda }=R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ ($R_H\approx 1.097\times 10^7{\text{ m}}^{-1}$)

Calculation:

\frac{1}{\lambda} = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{16}\right) = 1.097 \times 10^7 \left(\frac{3}{16}\right)$$$$\lambda \approx 486 \times 10^{-9} \text{ m}

Answer: $486\text{ nm}$ (Blue-green light)

Q11: Bohr Model (2018)

Question: Remove electron from $n=2$ (Ionization from first excited state).

Calculation:

$E_n=\frac{-13.6\text{ eV}}{n^2}\Rightarrow E_2=\frac{-13.6}{4}=-3.4\text{ eV}$

Energy required = $0-(-3.4)=3.4\text{ eV}$.

Answer: $3.4\text{ eV}$

4. Nuclear Physics

Q12: Radioactive Decay (2019, 21, 22)

Given: $T_{1/2}=4.51\times 10^9$ yrs. Time $t=10^{10}$ yrs.

Formula: $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

Calculation:

$\frac{N}{N_0}={\left(0.5\right)}^{10^{10}/4.51\times 10^9}=(0.5{)}^{2.217}\approx 0.215$

Answer: 21.5% survives.

Q13: Radioactive Decay (2020, 2024)

Given: $T_{1/2}=1600$ yrs. $N_0=1$ gm.

Calculation:

(i) Lose 1 centigram: Remaining $N=1-0.01=0.99$ gm.

$0.99=(1)e^{-\lambda t}\Rightarrow \ln (0.99)=-\frac{0.693}{1600}t\Rightarrow t\approx 23.2\text{ years}$

(ii) Reduced to 1 centigram: Remaining $N=0.01$ gm.

0.01 = (1) (0.5)^{t/1600} \Rightarrow \frac{t}{1600} = \frac{\ln(0.01)}{\ln(0.5)} \approx 6.64$$$$t = 6.64 \times 1600 \approx 10624 \text{ years}

Answer: (i) 23.2 years, (ii) 10,624 years

Q14: Activity (2017)

Given: $T_{1/2}=26.8$ min $=1608$ s. Activity $A=1$ Curie $=3.7\times 10^{10}$ decay/s.

Formula: $A=\lambda N\Rightarrow N=A/\lambda =A/(\ln 2/T_{1/2})$

Calculation:

$N=\frac{3.7\times 10^{10}\times 1608}{0.693}\approx 8.58\times 10^{13}\text{ atoms}$

Mass $m=\frac{N\times \text{Atomic Mass}}{N_A}=\frac{8.58\times 10^{13}\times 214}{6.023\times 10^{23}}\approx 3.05\times 10^{-8}\text{ g}$

Answer: $3.05\times 10^{-8}\text{ g}$

Q15: Decay to 60% (2018)

Given: $T_{1/2}=3.82$ days. Decay to 60% ($N=0.60N_0$).

Calculation:

0.60 = (0.5)^{t/3.82} \Rightarrow \frac{t}{3.82} \ln(0.5) = \ln(0.6)$$$$t = 3.82 \times \frac{-0.5108}{-0.693} \approx 2.81 \text{ days}

Answer: $2.81\text{ days}$

Q16: Nuclear Energy (2016)

Question: Mass defect and Binding Energy (BE) of Deuteron (${}_1^{2}H$).

Data (Standard): $m_p=1.007276$ u, $m_n=1.008665$ u, $m_D=2.014102$ u.

Calculation:

\Delta m = (m_p + m_n) - m_D = (1.007276 + 1.008665) - 2.014102$$$$\Delta m = 2.015941 - 2.014102 = 0.001839 \text{ u}$$$$BE = \Delta m \times 931.5 \text{ MeV} = 0.001839 \times 931.5 \approx 1.71 \text{ MeV}

(Note: If using standard text values where $\Delta m\approx 0.00238$ u, BE is $\approx 2.22$ MeV. The value depends on the precise masses used).

Answer: $\Delta m\approx 0.0024\text{ u}$, BE $\approx 2.22\text{ MeV}$

Q17: BE of Neutron in Li-7 (2023)

Interpretation: This asks for the Separation Energy of a neutron (${}_3^{7}Li{\to }_3^{6}Li+n$).

Data: $m{(}^{7}Li)\approx 7.01600$ u, $m{(}^{6}Li)\approx 6.01512$ u, $m_n\approx 1.00866$ u.

Calculation:

\Delta m = (m_{Li6} + m_n) - m_{Li7} = (6.01512 + 1.00866) - 7.01600$$$$\Delta m = 7.02378 - 7.01600 = 0.00778 \text{ u}$$$$E = 0.00778 \times 931.5 \text{ MeV} \approx 7.25 \text{ MeV}

Joules: $7.25\times 1.6\times 10^{-13}\text{ J}\approx 1.16\times 10^{-12}\text{ J}$.

Answer: $1.16\times 10^{-12}\text{ J}$ (or 7.25 MeV)

Q18: Fission Energy U-235 (2016)

Calculation:

Standard energy released per fission of U-235 is approximately 200 MeV.

Calculation involves: $\Delta m=M_{U235}+m_n-(M_{Ba}+M_{Kr}+3m_n)$.

$\Delta m\approx 0.215\text{ u}$.

$E=0.215\times 931.5\approx 200\text{ MeV}$.

Answer: $\approx 200\text{ MeV}$ (or $3.2\times 10^{-11}$ J)

Q19: Roppur Power Plant (2024)

Given: Power $P=250$ MW. Find rate of U-235 consumption per year.

Calculation:

Energy per fission $E=200\text{ MeV}=3.2\times 10^{-11}\text{ J}$.

Fissions per second $n=P/E=\frac{250\times 10^6}{3.2\times 10^{-11}}\approx 7.81\times 10^{18}\text{ fissions/s}$.

Mass per second = $\frac{n\times 235}{N_A}$ g.

Mass per year = $\frac{7.81\times 10^{18}\times 235}{6.023\times 10^{23}}\times (365\times 24\times 3600)$

= \frac{1835.35}{6.023 \times 10^{5}} \times 3.15 \times 10^7$$$$= 304 \times 10^{-5} \times 3.15 \times 10^7 \approx 96000 \text{ g} = 96 \text{ kg}

Answer: $\approx 96\text{ kg/year}$ (assuming 100% efficiency/thermal power).

Q20: Uncertainty Principle (2023)

Given: $\Delta t=10^{-8}$ s.

Formula: $\Delta E\cdot \Delta t\ge \frac{\hslash }{2}$ (or roughly $\approx \hslash$).

Using $\hslash \approx 1.054\times 10^{-34}\text{ J s}$.

Calculation:

$\Delta E\ge \frac{1.054\times 10^{-34}}{2\times 10^{-8}}\approx 0.527\times 10^{-26}\text{ J}$

In eV: $\frac{0.527\times 10^{-26}}{1.6\times 10^{-19}}\approx 3.29\times 10^{-8}\text{ eV}$.

Answer: $3.3\times 10^{-8}\text{ eV}$