The kinetic energy of an electron is E when the incident wavelength is `lamda` To increase ti KE of the electron to 2E, the incident wavelength must be

Options

A. `(hc)/(Elambda-hc)`
B. `(h lambda)/(Elambda+hc)`
C. `(hc lambda)/(Elambda+hc)`
D. `(hc lambda)/(Elambda-hc)`

Answer

c

Explanation

As `E=(hc)/lambda-phi_(0).......(i)`
and `2E=(hc)/(lambda')-phi_(0)........(ii)`
Subtracting (i) from (ii), we get
`2E=E=(hc)/(lambda')-(hc)/lambda or E+(hc)/lambda=(hc)/(lambda')`
or `lambda'=(hclambda)/(Elambda+hc)`