Question

The energy of the incident photon on the metal surface is 3 W and then 5 W, where W is the work function for that metal. The ratio of velocities of emitted photoelectrons is ______.

Options

• 1 : 4

• 1 : 2

• 1 : √2

• 1 : 1

Answer 1

The energy of the incident photon on the metal surface is 3 W and then 5 W, where W is the work function for that metal. The ratio of velocities of emitted photoelectrons is 1 : √2.

Explanation:

In photoelectric emission,
kinetic energy of emitted electron
= energy of incident photon - work-function

⇒ `1/2"mv"^2="E"-"W"`

or `"v"=sqrt((2("E"-"W"))/"m")`

∴ `"v"_1/"v"_2=sqrt((2("E"_1-"W"))/"m")xxsqrt("m"/(2("E"_2-"W")))=sqrt(("E"_1-"W")/("E"_2-"W"))`

`=sqrt((3"W"-"W")/(5"W"-"W"))=sqrt(2/4)=1/sqrt2 " or " 1 : sqrt2`