Single Choice

When a surface 1 cm thick is illuminated with light of wave length $$\lambda$$ the stopping potential is $$V_{0}$$ ,but when the same surface is illuminated by light of wavelength 3$$\lambda$$ , the stopping potential is $$\dfrac{V_{0}}{6}$$. The threshold wavelength for metallic surface is

A4$$\lambda$$
B5$$\lambda$$
Correct Answer
C3$$\lambda$$
D2$$\lambda$$

Solution

$$V_{0}\ =\ \dfrac{hc}{e}\left ( \dfrac{1}{\lambda }-\dfrac{1}{\lambda_{0}} \right )$$ ( l )

$$\ \frac{V_{0}}{6}=\dfrac{hc}{e}\left ( \dfrac{1}{3\lambda }-\dfrac{1}{\lambda_{0}} \right )$$ (ll)

Now equation ( I ) $$\div $$ equation ( l l )

$$6=\ \dfrac{3(\lambda _{0}-\lambda )}{\lambda _{0}-3\lambda }$$

$$\ 6\lambda _{0}-18\lambda \ =3\lambda _{0}-3\lambda$$
$$\lambda _{0}=5\lambda$$
So, the answer is option (B).

SIMILAR QUESTIONS

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