When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is V…

Question

When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is V4 . If the threshold wavelength for the metallic surface is nλ then value of n will be:

Accepted Answer

Correct answer: 9. (9) hc ⋋- = eV … . . (i ) hc3 ⋋- = eV4 … ( ii ) From (i) and (ii) hc3 ⋋- = hc4 ⋋- 4 ⇒ hc ⋋( 13- 14)= 3 4 ⇒ hc ⋋ 19= Now if = hc ⋋_0 ∴ ⋋_0=9 ⋋ So, n=9

$When radiation of wavelength λ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength 3λ, the stopping potential is \frac{V}{4} . If the threshold wavelength for the metallic surface is nλ then value of n will be:$

Solution

More Atoms PYQs