Question

​​Determine maximum kinetic energy of electron emitted by silver surface when illuminatedby light of wavelength 2500 Å. The threshold wavelength of silver is 2762 Å. (h =6. 61x10 ​-34J.sec. and c=3x10​8​ m/sec).

Answer 1

Solution

Maximum kinetic energy is given by

$K_{max}=\frac{hc(\lambda_0-\lambda)}{\lambda_0\lambda}$

Putting all values

$K_{max}=\frac{6.61\times10^{-34}\times3\times10^8(2762-2500) 10^{-10} }{2762\times 2500\times10^{-20}}\\K_{max}= 7.5\times10^{-20} Joule$