The K.E. of the electron is E, when the incident wavelength is `lamda`. To increase the K.E. of the electron to 2E, the incident wavelength must be :

Options

A. `(hc)/(Elamda-hc)`
B. `(hclamda)/(Elamda+hc)`
C. `(hlamda)/(Elamda+hc)`
D. `(hclamda)/(Elamda-hc)`

Answer

B

Explanation

`(hc)/(lambda)=hv_(0)+E and (hc)/(x)=hv_(0)+2E`
`or (hc)/(lambda)-E=(hc)/(x)-2E,` Solving `x=(hc lambda)/(lambda+hc)`