Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons are

5.5 V

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

1.5 V

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

9.5 V

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

4.0 V

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

Open in App

Solution

The correct option is D
4.0 V

$K_{m a x} = e V_{0} \Rightarrow e V_{0} = 4 e V \Rightarrow V_{0} = 4 V$

Suggest Corrections

Similar questions

Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons are

Q. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy

6

eV fall on it is

4

eV. The stopping potential is?

Q. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy

6 e V

fall on it is

4 e V

. The stopping potential in volt is :

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is

Join BYJU'S Learning Program

Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons are

The correct option is D 4.0 V Kmax=eV0⇒eV0=4eV⇒V0=4V

The correct option is D
4.0 V

$K_{m a x} = e V_{0} \Rightarrow e V_{0} = 4 e V \Rightarrow V_{0} = 4 V$