Question

The maximum velocity of the photoelectron emitted by the metal surface is v. Charge and the mass of the photoelectron is denoted by e and m, respectively. The stopping potential in volt is ______.

Options

• `"v"^2/(2("e"/"m"))`

• `"v"^2/(("m"/"e"))`

• `"v"^2/(2("m"/"e"))`

• `"v"^2/(("e"/"m"))`

Answer 1

The maximum velocity of the photoelectron emitted by the metal surface is v. Charge and the mass of the photoelectron is denoted by e and m, respectively. The stopping potential in volt is `"v"^2/(2("e"/"m"))`.

Explanation:

The maximum velocity of emitted photoelectron from the metal surface at stopping potential is given by

`"K"_"max" = 1/2"mv"^2="eV"_0`

∴ Stopping potential, `"V"_0="mv"^2/(2"e")="v"^2/(2("e"/"m"))`