Question

Light with a wavelength of 395 nm illuminates a metal cathode. The maximum kinetic energy of the emitted electrons is 0.76 eV . Part A What is the longest wavelength of light that will cause electrons to be emitted from this cathode

Answer 1

Answer:

520.8 nm

Explanation:

We are given that

$\lambda=395 nm=395\times 10^{-9} m$

1 nm=$10^{-9} m$

Maximum kinetic energy,$K_{max}=0.76 eV=0.76\times 1.6\times 10^{-19}$ V

$1 e=1.6\times 10^{-19} C$

We have to find the maximum wavelength of light.

We know that

$h\nu=\frac{hc}{\lambda_0}+K_{max}$

Where $c=3\times 10^8 m/s$

$h=6.625\times 10^{-34}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}+0.76\times 1.6\times10^{-19}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\lambda_0=\frac{6.625\times 10^{-34}\times 3\times 10^8}{6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}}$

$\lambda_0=5.208\times 10^{-7} m$=520.8 nm