Question

The wavelength of light incident on a metal surface is reduced from 300 nm to 200 nm (both are less than threshold wavelength). What is the change in the stopping potential for photoelectrons emitted from the surface will be ______ V. (Take h = 6.6 × 10^-34 J-s)

Options

• 1

• 2

• 3

• 4

Answer 1

The wavelength of light incident on a metal surface is reduced from 300 nm to 200 nm (both are less than threshold wavelength). What is the change in the stopping potential for photoelectrons emitted from the surface will be 2 V. (Take h = 6.6 × 10^-34 J-s)

Explanation:

`"eV"_1 = "hc"/lambda_1-phi`

`"eV"_2 = "hc"/lambda_2-phi`

`"e"("V"_2-"V"_1) = "hc"((lambda_1-lambda_2)/(lambda_1lambda_2))`

`"V"_2 - "V"_1 = "hc"/"e"((lambda_1-lambda_2)/(lambda_1lambda_2))`

= `(6.6xx10^-34xx3xx10^8)/(1.6xx10^-19)xx100/(6xx10^-5)`

= `66/32xx10^(-34+8+2+19+5)`

= `33/16`

= 2.0625 volt ≈ 2 volt