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When a surface is irradiated with light of wavelength 4950 $\overset{0}{A}$ , a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. The wavelength of the second source is

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5125 $\overset{0}{A}$

3150 $\overset{0}{A}$

4133 $\overset{0}{A}$

4523 $\overset{0}{A}$

answer is B.

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Detailed Solution

in the first case enrgy of the photon is $\frac{12400}{4950} \approx 2.5 e V$ and maximum kinetic energy of the electron is 0.6 eV so form Einstein equation $h ν = W + e V_{s}$

2.5=W+0.6 W=1.9 eV

for the second case $h ν = W + e V_{s}; h ν = 1.9 + 1.1 = 3 e V$

corresponding wavelength $\frac{12400}{3} = 4133$

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