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 10¹⁴ 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 10¹⁴ 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 10⁸ 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