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qaws [65]
3 years ago
5

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.
Physics
1 answer:
poizon [28]3 years ago
5 0

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

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