JEE Physics Dual Nature Of Radiation And Matter MCQs Set 16

Answer: (b) \( 4.2\times10^{-25}\text{ kg m/s} \)

Explanation:
1. Photon energy \( \dfrac{12400}{4000}=3.1\text{ eV} \), so \( K_{max}=3.1-2.5=0.6\text{ eV} \).
2. \( p=\sqrt{2mK}=\sqrt{2\times9.1\times10^{-31}\times0.6\times1.6\times10^{-19}}\approx4.2\times10^{-25}\text{ kg m/s} \).

Teacher's Note:
1. Convert the kinetic energy to joules before using the momentum formula.
2. The momentum is for the fastest electrons only.

Answer: (d) a decrease of 0.054 V

Explanation:
1. \( eV_0=\dfrac{hc}{\lambda}-\phi \), so \( \Delta V_0=12400\left(\dfrac1{3040}-\dfrac1{3000}\right)=-0.054\text{ V} \).

Teacher's Note:
1. Longer wavelength means lower photon energy and lower stopping potential.
2. The work function cancels in the difference.

Answer: (c) \( 3.2\ \mu\text{A} \)

Explanation:
1. Photon energy \( \dfrac{1240}{800}=1.55\text{ eV} \), so photons per second \( \dfrac{5}{1.55\times1.6\times10^{-19}}=2.0\times10^{19} \).
2. Electrons per second \( 2.0\times10^{13} \), current \( =2.0\times10^{13}\times1.6\times10^{-19}=3.2\times10^{-6}\text{ A} \).

Teacher's Note:
1. Saturation current counts every emitted electron.
2. The photocurrent is proportional to the intensity at a fixed frequency.

Answer: (a) \( 5\times10^{-8}\text{ N} \)

Explanation:
1. Momentum of a photon \( p=\dfrac h\lambda=\dfrac{6.63\times10^{-34}}{663\times10^{-9}}=10^{-27}\text{ kg m/s} \).
2. For reflection, the change in momentum along the normal is \( 2p\cos60^\circ \) per photon, so \( F=5\times10^{19}\times2\times10^{-27}\times0.5=5\times10^{-8}\text{ N} \).

Teacher's Note:
1. Only the component of momentum normal to the mirror changes.
2. The angle of incidence is measured from the normal.

Answer: (d) \( 1.3\times10^{-7}\text{ N} \)

Explanation:
1. Momentum delivered per second: absorbed part \( 0.7\dfrac Pc \) and reflected part \( 0.3\times\dfrac{2P}{c} \).
2. Force \( =\dfrac Pc(0.7+0.6)=\dfrac{30}{3\times10^8}\times1.3=1.3\times10^{-7}\text{ N} \).

Teacher's Note:
1. A perfect absorber receives P/c; a perfect reflector receives 2P/c at normal incidence.
2. The force is the total momentum transferred per second.

Answer: (b) 12 Å

Explanation:
1. \( K_{max}=12400\left(\dfrac1{3000}-\dfrac1{4000}\right)=1.03\text{ eV} \).
2. For an electron \( \lambda=\dfrac{12.27}{\sqrt{K(\text{eV})}}\text{ Å}=\dfrac{12.27}{\sqrt{1.03}}\approx12\text{ Å} \).

Teacher's Note:
1. For an electron accelerated through V volts, lambda = 12.27/sqrt(V) angstrom.
2. Maximum kinetic energy corresponds to the threshold subtraction.

Answer: (a) becomes more than double

Explanation:
1. \( K_{max}=h\nu-\phi \). Doubling \( \nu \) gives \( K'=2h\nu-\phi=2K_{max}+\phi \), which is more than twice \( K_{max} \).

Teacher's Note:
1. The work function is subtracted once, not twice.
2. Kinetic energy is linear in frequency but does not pass through the origin.

Answer: (c) \( 4\lambda \)

Explanation:
1. \( eV=\dfrac{hc}\lambda-\phi \) and \( \dfrac{eV}{3}=\dfrac{hc}{2\lambda}-\phi \).
2. Eliminating \( eV \): \( \dfrac{hc}{\lambda}-\phi=\dfrac{3hc}{2\lambda}-3\phi\Rightarrow2\phi=\dfrac{hc}{2\lambda} \), so \( \phi=\dfrac{hc}{4\lambda} \).
3. Threshold wavelength \( \lambda_0=\dfrac{hc}{\phi}=4\lambda \).

Teacher's Note:
1. Write the photoelectric equation twice and eliminate the stopping potential.
2. The threshold wavelength corresponds to zero kinetic energy.

Answer: (b) \( 1.5\times10^{12} \)

Explanation:
1. Power on 1 cm² \( =1\times10^{-4}\text{ W} \). Photon energy \( \dfrac{1240}{300}=4.13\text{ eV}=6.6\times10^{-19}\text{ J} \).
2. Photons per second \( =1.5\times10^{14} \); 1% gives \( 1.5\times10^{12} \).

Teacher's Note:
1. Convert the area to square metres.
2. Only 1% of the photons eject electrons.

Answer: (d) \( \dfrac14 \)

Explanation:
1. Equal wavelengths mean equal momentum \( p \).
2. Electron: \( K=\dfrac{p^2}{2m}=\dfrac{pv}{2} \). Photon: \( E=pc \).
3. Ratio \( \dfrac{K}{E}=\dfrac{v}{2c}=\dfrac{1.5\times10^8}{6\times10^8}=\dfrac14 \).

Teacher's Note:
1. Photon energy is momentum times c.
2. The electron speed is half of c here.

Answer: (c) 1 : 1

Explanation:
1. The initial momentum is zero, so the two fragments have equal and opposite momenta.
2. Since \( \lambda=\dfrac hp \), the wavelengths are equal.

Teacher's Note:
1. Momentum is conserved in the decay.
2. Only the magnitude of the momentum enters the wavelength.

Answer: (a) \( \dfrac{h}{\sqrt{2meV}} \)

Explanation:
1. Kinetic energy \( \dfrac{p^2}{2m}=eV \), so \( p=\sqrt{2meV} \).
2. \( \lambda=\dfrac hp=\dfrac{h}{\sqrt{2meV}} \).

Teacher's Note:
1. For electrons this gives 12.27 over root V angstrom.
2. Wavelength decreases as the accelerating voltage increases.

Answer: (d) 3

Explanation:
1. \( \lambda=\dfrac{h}{\sqrt{2mqV}} \), so \( \dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt{4\times2}=\sqrt8\approx3 \).

Teacher's Note:
1. The alpha particle has four times the mass and twice the charge of the proton.
2. The heavier, more highly charged particle has the shorter wavelength.

Answer: (b) 1

Explanation:
1. \( V_0=\dfrac{h}{e}\nu-\dfrac{\phi}{e} \): the slope is \( h/e \) for every metal, so the ratio is 1.

Teacher's Note:
1. The work function only shifts the intercept.
2. This slope was used by Millikan to measure Planck's constant.

Answer: (a) 3.7 eV

Explanation:
1. Photon energies: \( \dfrac{1240}{248}=5.0\text{ eV} \) and \( \dfrac{1240}{310}=4.0\text{ eV} \).
2. Kinetic energies are in the ratio 4 : 1: \( 5-\phi=4(4-\phi)\Rightarrow3\phi=11 \), so \( \phi=3.67\text{ eV} \).

Teacher's Note:
1. Kinetic energy goes as the square of the speed.
2. Solve the linear equation for the work function.

Answer: (c) Barium and lithium; 0.46 eV (lithium)

Explanation:
1. Photon energy \( \dfrac{1240}{450}=2.76\text{ eV} \).
2. Ejection needs \( \phi<2.76\text{ eV} \): barium (2.5) and lithium (2.3).
3. The lowest work function gives the largest energy: \( 2.76-2.3=0.46\text{ eV} \) for lithium, which is also the largest stopping potential, 0.46 V.

Teacher's Note:
1. The metal with the smallest work function gives the fastest electrons.
2. Tantalum and tungsten need photons above 4 eV.

Answer: (b) 151 eV

Explanation:
1. Nodes at atoms mean \( d=n\dfrac\lambda2 \). Consecutive spacings \( n\dfrac\lambda2=2 \) and \( (n+1)\dfrac\lambda2=2.5 \) give \( \dfrac\lambda2=0.5\text{ Å} \), \( n=4 \) and \( \lambda=1\text{ Å} \).
2. Energy \( =\dfrac{12.27^2}{\lambda^2}=150.6\approx151\text{ eV} \).

Teacher's Note:
1. Subtract the two conditions to find half a wavelength.
2. The least spacing allowing such a wave is half a wavelength, 0.5 Å.

Answer: (d) the stopping potential stays 0.6 V and the saturation current is 2.0 mA

Explanation:
1. The stopping potential depends only on the frequency and the metal, not on the distance, so it stays 0.6 V.
2. Intensity falls as \( \dfrac1{d^2} \): the factor is \( \dfrac1{9} \), so the saturation current is \( \dfrac{18}{9}=2.0\text{ mA} \).

Teacher's Note:
1. Distance changes the intensity and hence the number of photons, not their energy.
2. The inverse square law applies to a point source.

Answer: (c) 2.0 V and \( 3.2\times10^{-19}\text{ J} \)

Explanation:
1. \( K_{max}=5-3=2\text{ eV} \), so the stopping potential is 2.0 V.
2. In joules \( 2\times1.6\times10^{-19}=3.2\times10^{-19}\text{ J} \).

Teacher's Note:
1. The stopping potential in volts equals the kinetic energy in eV.
2. Convert eV to joules by multiplying by 1.6e-19.