Question

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 3.35 × 1014 s–1.

With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of λ = 265 nm?

Answer #1

1) Given, = 3.35 x
10^{14} s^{-1}

We have the relation as-- E = h , where h is the
Planck's constant (6.626 x 10^{-34} Js)

Now , E = h = 6.626 x
10^{-34} Js * 3.35 x 10^{14} s^{-1} =
22.1971 x 10^{-20} J = 2.2 x 10^{-19} J per
photon

2) Given , = 265 nm = 265
x 10^{-9} m

We have the relation as, E = hc /

E=( 6.626 x10^{-34} Js * 3 x 10^{8} m/s )/ 265 x
10^{-9} m

E = 19.878 x 10^{-26} / 265 x 10^{-9} J

E = (19.878/265 ) x 10^{-17} J

E= 7.5 x 10^{-2} x 10^{-17} J

E = 7.5 x 10^{-19} J per photon.

We know,

hν = W + E(k) where,

W = the minimum energy required to remove an electron from the
surface of the metal

E(k) = kinetic energy of the ejected electron

E(k) =( 7.5 x 10^{-19} J) - ( 2.2 x 10^{-19}
J)

E(k) = (7.5 -2.2) x 10^{-19} J = **5.3 x
10 ^{-19} J**

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