# NEET Questions Solved

NEET - 2016

When a metallic surface is illuminated with radition of wavelength $\lambda$, the stopping potential is V, if the same surface is liiuminated with radiation of wavelength 2$\lambda$, the stopping potential is $\frac{V}{4}$ The threshold wavelength for matallic surface is

(a) 5$\lambda$                (b)$\frac{5}{2}$$\lambda$

(c) 3$\lambda$                (d) 4$\lambda$

(c) In Ist case, when a matallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V.

So, photoelectric equation can be written as

eV=$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_{ο}}$                        ...(i)

In IInd Case, when the same surface is illuminayed with radition of wavelength 2$\lambda$, the sopping potential is $\frac{V}{4}$ so, photoelectric equation can be written as

$\frac{eV}{4}=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_{0}}$

...(ii)

From eqs, (i) and (ii) we get

Difficulty Level:

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