Photoelectrons are emitted from a metal surface when photons of energy 3.1eV or more impinge on it .In an experiment with this metal using a monochromatic radiation the stopping potenital was found to be 2.5V. The energy of the photons used in the experment is

Options

A. 7.8eV
B. 6.2eV
C. 5.6eV
D. 4.2eV

Answer

C

Explanation

Energy of photons,E=3.1eV
Stopping potential ,`V_(o)`=2.5V
The energy of photons can be calculated from Einsteins photoelectric equation .
`implies (KE)_(max)`=E-W (1)
(i)Initially when K.E is zero
0=E-W
`impliesW=E`
`thereforeW=3.1eV`
(ii) When stopping potential is appiled .
`(K.E)_(max)=eV_(o)` `implies E=(K.E)_(max)+W`(because From equation (1))
E=2.5+3.1
`therefore`E=5.6eV