Question

When certain metal surface is illuminated with a light of wavelength A., the stopping potential is V, When the same surface is illuminated by light of wavelength 2λ, the stopping potential is `("V"/3)`. The threshold wavelength for the surface is ______.

Options

• `(8lambda)/3`

• `(4lambda)/3`

• 4λ

• 6λ

Answer 1

When certain metal surface is illuminated with a light of wavelength A., the stopping potential is V, When the same surface is illuminated by light of wavelength 2λ, the stopping potential is `("V"/3)`. The threshold wavelength for the surface is 4λ.

Explanation:

Given for a metal, wavelength of light used = λ

Stopping potential = V

If λ₀ be the threshold wavelength, then maximum kinetic energy of emitted electrons

`"K"_"max" = "hc"(1/lambda - 1/lambda_0)`    ...(i)

Again, wavelength of used light

λ' = 2λ

Stopping potential, V' = `"V"/3`

then `"K"_"max" = "hc"(1/(lambda') - 1/lambda_0)`

`=> "eV'" = "hc"(1/(2lambda) - 1/lambda_0)`

`=> "eV"/3 = "hc"(1/(2lambda) - 1/lambda_0)`   ...(ii)

From Eqs. (i) and (ii), we have

`"eV"/("eV"/3) = ("hc"(1/lambda - 1/lambda_0))/("hc"(1/(2lambda) - 1/lambda_0))`

`3 (1/(2lambda) - 1/lambda_0) = 1/lambda - 1/lambda_0`

`=> lambda_0 = 4lambda`

So, threshold wavelength is 4 times of wavelength of light.