Light of wavelength 330nm falling on a piece of metal ejects electrons with sufficient energy with required voltage `V_0` to prevent them reaching a collector. In the same set up, light of wavelength 220 nm ejects electrons which require twice the voltage `V_0` to stop them in reaching a colleator. the numerical value of voltage `V_0` is

Options

A. `(16)/(15) V`
B. `(15)/(16) V`
C. `(15)/(8) V`
D. `(8)/(15) V`

Answer

C

Explanation

`ev_0 =(hc)/(lambda_1) -W`
Similarly, e(2V_0) = (hc)/(lambda^2) -W`
Subtracting Eq. (i) from Eq (ii) we get
`eV_0 = (hc(lambda_1 - lambda_2))/(lambda_1lambda_2)`
`:. V_0 =(hc(lambda_1-lambda_2))/(elambda_1lambda_2)`
`=(6.63xx10^(-34))(3xx10^8)(110))/(1.6xx 10^(-19)xx330xx220xx10^(-9))`
`~~ 1.8 volt`