Question

The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7 × 105 m/s when the wavelength of the light is λA. However, a maximum speed of vB = 5.1 × 105 m/s is observed when the wavelength is λB. Find the wavelengths λA and λB.

Answer 1

Answer:

$\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=3.32344\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.63\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

$W_0$ = Work function = $4.8\times 10^{-19}\ J$

$v_A$ = Velocity of A particle = $7.7\times 10^5\ m/s$

$v_B$ = Velocity of B particle = $5.1\times 10^5\ m/s$

The wavelength is given by

$\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}$

$\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m$

The wavelength $\lambda_A=2.65177\times 10^{-7}\ m$

$\lambda_B=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(5.1\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_B=3.32344\times 10^{-7}\ m$

The wavelength $\lambda_B=3.32344\times 10^{-7}\ m$