Initially a photon of wavelength `lambda_(1)`falls on the photocathode and emits of maximum energy `E_(1)`.If the wavelength of the incident photon is changed to `lambda_(2)` the maximum energy of the electron emitted becomes` E_(2)`.Then value of he (h=Planck's constant,c= velocity of light) is

Options

A. `hc=((E_(1)+E_(2))(lambda_(1)lambda_(2)))/(lamda_(2)-lamda_(2)`
B. `hc=(E_(1)-E_(2))/(lamda_(2)lamda_(1)),(lamda_(1)lamda_(2))`
C. `hc= ((E_(1)-E_(2))(lamda_(2)lamda_(1)))/(lamda_(1)lamda_(2)`
D. `hc=(lamda_(2)-lamda_(1))/(lamda_(1)lamda_(2))E1`

Answer

B

Explanation

From the equation of photoelectric effect,
`E_(1)=(hc)/lamda_(1)-W`
`impliesE_(1)+W=(hc)/(lamda_(1))`
`E_(2)=(hc)/(lamda_(2))-WimpliesE_(2)+W=(hc)/(lamda_(2))`
From equation
`W=(hc)/(lamda_(2))-E_(2)`
Subsituting the value of W in equation (1),
`E_(1)+(hc)/(lamda(2))-E_(2)=(hc)/(lamda_(1)`
`E_(1)-E_(2)=hc[(1)/(lamda_(1))-(1)/(lamda_(2))]`
`=hc((lamda_(2)-lamda_(1))/(lamda_(1)lamda_(2)))therefore hc= ((E_1-E_(2))(lamda_(2)-lamda_(1)))/((lamda_(1)-lamda_(2))`