A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are `mu_1` and `mu_2` respectively. If the ratio `u_1: u_2 = 2:1 and hc= 1240 eV` , the work function of the metal is nearly.
(a)3.7 eV (b) 3.2 eV
(c ) 2.8eV (d) 2.5eV.

Options

A. 3.7 eV
B. 3.2 eV
C. 2.8 eV
D. 2.5 eV

Answer

A

Explanation

`(hC)/(lamda_1) -W = 1/2 mu_1^2`
and `(hC)/(lamda_2) -W = 1/2 mu_2^2`
dividing the above two equations , we get
`( (hC)/(lamda_1) - W )/( (hC)/(lamda_2) - W) = (u_1^2)/(u_2^2) therefore (1240/248 - W)/(1240/310 - W) = 4/1`
` therefore 1240/248 - W = (4 xx 1240)/(310 ) - 4W therefore W = 3.7 eV`