Question

The light of wavelength '`lambda`'. incident on the surface of metal having work function `phi` emits the electrons. The maximum velocity of electrons emitted is ______.
[c = velocity of light, h = Planck's constant, m = mass of electron]

Options

• `[(2("hv" - phi)lambda)/"mc"]`

• `[(2("hc" - lambdaphi))/("m"lambda)]^(1/2)`

• `[(2("hc" - lambda))/("m"lambda)]^(1/2)`

• `[(2("hc" - phi))/("m"lambda)]`

Answer 1

The light of wavelength '`lambda`'. incident on the surface of metal having work function emits the electrons. The maximum velocity of electrons emitted is `underline([(2("hc" - lambdaphi))/("m"lambda)]^(1//2))`

Explanation:

In photoelectric effect, the maximum kinetic energy possessed by the particle,

`"KE"_"max" = "hv" - phi`

`1/2 "mv"^2 = "hc"/lambda - phi   ...(because "v" = "c"/lambda)`

`=> "v"^2 = (2("hc" - philambda))/(lambda"m")`

v = `[(2("hc" - lambdaphi))/("m"lambda)]^(1//2)`