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Find the velocity of electrons liberated by electromagnetic radiation of 18 nm from stationary He+ ions in the ground state.

goo , 7 Years ago
Grade 6
anser 1 Answers
Arun
the work function W = 13.6 x 4 eV = 54, 4 eV
​Energy of the falling photon = h c / e L
h = 6.626 x 10^-34 J s
c = 3 x 10^8 m/s
e = 1.6 x 10^-19 C
L = wavelength lambda = 18 x 10^-9 m
So energy of photon computed to be 69.02 e V
Hence energy of photo electron = 69.02 - 54.4 = 14.62 e V
This equals to kinetic energy of photo electron 1/2 * m * v^2
Hence v = ( 2 x 14.62 x 1.6 x 10^-19 / 9 x 10-31)^1/2
By calculator v = 2.28 x 10^6 m/s
Last Activity: 7 Years ago
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