When a photon of energy $7 e V$ is made incident on a metal then the emitted electron is stopped by a stopping potential of $- 5.5 V$ . The work function of metal will be :

- 1.5 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 e V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

12.5 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

37.5 e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1.5 e V$
The stopping potential $(V_{s})$ related to the incident light energy (E) by the following relation:

$e V_{s} = E - ϕ$ .

So, the work-function is given by:

$ϕ = E - e V_{s} = (7 - 5.5) = 1.5 e V$

So, the answer is option (B).

Suggest Corrections

Similar questions

Q. A photon of

6 e V

is incident on a metal surface and the emitted electron is stopped by a retarding potential of

- 5 e V

. What is the work function of the metal?

Q. A source of light is placed above a sphere of radius

10 c m

. How many photoelectrons must be emitted by the sphere before emission of photoelectrons stops? The energy of incident photon is

4.2 e V

and the work function of the metal is

1.5 e V

Q. In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of 1.2 V. Find energy of incident photon.

When an electromagnetic radiation of wavelength $310 n m$ falls on the surface of sodium metal, electrons are emitted with $K . E_{m a x} = 1.5 e V$ . Determine the work function $W_{o}$ of metal.

Q. A photon of energy

4 e V

is incident on a metal surface whose work function is

2 e V

. The minimum reverse potential to be applied for stopping the emission of electrons is-

Join BYJU'S Learning Program

When a photon of energy 7 eV is made incident on a metal then the emitted electron is stopped by a stopping potential of −5.5 V. The work function of metal will be :

The correct option is B 1.5eVThe stopping potential (Vs) related to the incident light energy (E) by the following relation:eVs=E−ϕ.So, the work-function is given by:ϕ=E−eVs=(7−5.5)=1.5eVSo, the answer is option (B).

When a photon of energy $7 e V$ is made incident on a metal then the emitted electron is stopped by a stopping potential of $- 5.5 V$ . The work function of metal will be :

The correct option is B $1.5 e V$
The stopping potential $(V_{s})$ related to the incident light energy (E) by the following relation:

$e V_{s} = E - ϕ$ .

So, the work-function is given by:

$ϕ = E - e V_{s} = (7 - 5.5) = 1.5 e V$

So, the answer is option (B).