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When light with a wavelength of 262 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.47 x 10^-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
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Added Tue, 21 Jul '15
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solution:- Planck's Constant,h = (6.63*10^-34)J-s ; velocity of light ,v = (3*10^8)m/s ; wavelength is denoted by 'L'

now, Incident energy = Work function + K.E

energy = h*c/L

Now, K.E = Incident energy - Work function

Thus, as per question:-

3.47*10^-19 J = {(6.63*10^-34)*(3*10^8)/(262*10^-9)} -  Work function...........(1)

and 6.94*10^-19 J = {(6.63*10^-34)*(3*10^8)/(L)} -  Work function...........(2)

(2) - (1) ; we get

3.47*10^-19 J = {(6.63*10^-34)*(3*10^8)}*[(1/L) - (1/262*10^-9)]

or, L = 1.798*10^-7 m = 179.8 nm
Edit
Added Tue, 21 Jul '15
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