JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 26)
A certain metallic surface is illuminated by monochromatic radiation of wavelength $$\lambda$$. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2$$\lambda$$, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ____________ $$\lambda$$.
Answer
4
Explanation
$$KE = {{hc} \over \lambda } - \phi hc$$
$$e(3{V_0}) = {{hc} \over \lambda } - \phi $$ ..... (i)
$$e{V_0} = {{hc} \over {2\lambda }} - \phi $$ ..... (ii)
Using (i) & (ii)
$$\phi = {{hc} \over {4{\lambda _0}}} = {{hc} \over {{\lambda _t}}}$$
$${\lambda _t} = 4{\lambda _0}$$
$$e(3{V_0}) = {{hc} \over \lambda } - \phi $$ ..... (i)
$$e{V_0} = {{hc} \over {2\lambda }} - \phi $$ ..... (ii)
Using (i) & (ii)
$$\phi = {{hc} \over {4{\lambda _0}}} = {{hc} \over {{\lambda _t}}}$$
$${\lambda _t} = 4{\lambda _0}$$
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