JEE Physics Dual Nature Of Radiation And Matter MCQs Set 08

Answer: (d) Graph (d)

Explanation:
1. Einstein's equation gives \(K_{max}=h\nu-\phi\)
2. For \(\nu<\nu_0\) no electron is emitted, so the graph is flat at zero up to the threshold frequency
3. Beyond the threshold the graph is a straight line of slope \(h\)
4. Only graph (d) shows this; (b) is wrong because K cannot be positive at zero frequency and (c) ignores the threshold

Teacher's Note:
1. The intercept of the straight part on the frequency axis is the threshold frequency and the slope is Planck's constant for every metal. Remember the graph has no point on the vertical axis for positive kinetic energy.

Answer: (b) greater than \(\sqrt2\,v\)

Explanation:
1. \(\tfrac12mv^2=h\nu-\phi\)
2. At \(2\nu\): \(\tfrac12mv'^2=2h\nu-\phi=2(h\nu-\phi)+\phi\)
3. Since \(\phi>0\), \(v'^2>2v^2\), so \(v'>\sqrt2\,v\)

Teacher's Note:
1. Doubling the frequency more than doubles the kinetic energy because the work function does not change. Speed is the square root of kinetic energy, so it grows by more than \(\sqrt2\).

Answer: (a) \(2hc\times10^{6}\ \text{J}\)

Explanation:
1. \(\frac{hc}{\lambda_1}-\phi=\tfrac12mv^2\) and \(\frac{hc}{\lambda_2}-\phi=4\cdot\tfrac12mv^2\)
2. Subtracting: \(hc\left(\frac1{250}-\frac1{400}\right)\times10^{9}=3K\), so \(K=0.5\,hc\times10^{6}\) J
3. \(\phi=\frac{hc}{400\times10^{-9}}-K=2.5hc\times10^6-0.5hc\times10^6=2hc\times10^6\) J

Teacher's Note:
1. Write the photoelectric equation for both wavelengths and eliminate the unknown kinetic energy. Keep wavelengths in metres so the powers of ten stay correct.

Answer: (b) the kinetic energy of the electrons reaching the collector will increase

Explanation:
1. A downward field exerts an upward force on the (negative) electrons, directed towards the collector
2. The electrons are accelerated, so they reach the collector with greater kinetic energy
3. The current was already saturated, so it cannot increase; the stopping potential and threshold wavelength depend only on the light and the metal

Teacher's Note:
1. In saturation every emitted electron already reaches the collector. A field only changes the energy with which they arrive, not the number emitted.

Answer: (c) \(2\sqrt2:1\)

Explanation:
1. \(\lambda=\frac{h}{\sqrt{2mqV}}\)
2. \(\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha q_\alpha}{m_pq_p}}=\sqrt{\frac{4\times2}{1\times1}}=2\sqrt2\)
3. So the ratio is \(2\sqrt2:1\)

Teacher's Note:
1. For the same accelerating voltage the lighter and less charged particle has the longer wavelength. Use \(m_\alpha=4m_p\) and \(q_\alpha=2q_p\).

Answer: (d) Graph (d)

Explanation:
1. de-Broglie relation: \(p=\frac h\lambda\)
2. So \(p\) is inversely proportional to \(\lambda\)
3. This is a rectangular hyperbola, shown in graph (d)

Teacher's Note:
1. An inverse relationship always gives a hyperbola with both axes as asymptotes. A straight line would mean direct proportionality.