A metallic surface is illuminated alternatively with lights of wavelengths `3000A` and `6000A`. It is obseved that the maximum speeds of the photoelectrons under these illumination are in the ratio `3:1`
Q. The work function of the metal is

Options

A. 1.45 eV
B. 2.26 eV
C. 1.23 eV
D. 3.4 eV

Answer

C

Explanation

From Einstein's photoelectric effect equation,
`(hc)/(lamda)=phi_0+(1)/(2)mv^2`
For `lamda_1=3000A`,
`(hc)/(3000xx10^(-10))=phi+(1)/(2)m(3v)^2` ...(i)
For `lamda_1=6000A`,
`(hc)/(6000xx10^(-10))=phi_0+(1)/(2)mv^2` ...(ii)
Now, on multiplying Eq. (ii) by 9 and subracting Eq. (i) from it, we get
`8phi_0=9(hc)/(6000xx10^(-10))-(hc)/(3000xx10^(-10))`
Substituting the values, we get
`phi_0=1.23 eV`
Maximum speed of the photoelectron will be for the incident light of wavelength `lamda=3000A`. From Eq. (i)
`(1)/(2)m(3v)^2=(6.62xx10^(-34)xx3xx10^(8))/(3000xx10^(-10))-2.896xx10^(-19)`
`=3.724xx10^(-19)`
`3v=v_(max)(((2xx3.724xx10^(-19)))/(9.1xx10^(-31)))^((1)/(2))=9xx10^(5)ms^(-1)`