Light of wavelength $200 n m$ shines on an aluminium, $4.20 e V$ is required to eject an electron. What is the kinetic energy of the fastest ejected electrons?

0.5 e V

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

1.00 e V

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

2.00 e V

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

4.00 e V

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

Open in App

Solution

The correct option is C $2.00 e V$
Using the Einstein's equation of photoelectric effect,
$\frac{h c}{λ} = K E + ϕ$
$K E = \frac{1240}{200} - 4.2$
$K E = 2 e V$

Suggest Corrections

Similar questions

Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength $2000 \circ A$ falls on the surface, the velocity of the fastest electron ejected from the surface will be

Q. Light of wavelength 200

n m

falls on an aluminium plate. If work function of plate is 2

e V

, then kinetic energy of the fastest electron could be

Q. The minimum energy required by an electron to overcome the attractive forces of the nucleus is

0.6 \times 10^{- 19} J

. What is the maximum kinetic energy of the ejected electron if the material is radiated with a light of wavelength

2210 nm

Q. The work function of tungsten is

4.50 e V

. Calculate the speed of fastest electron ejected from tungsten surface when light whose photon energy is

5.80 e V

shines on the surface.

Join BYJU'S Learning Program

Light of wavelength 200nm shines on an aluminium, 4.20eV is required to eject an electron. What is the kinetic energy of the fastest ejected electrons?

The correct option is C 2.00eVUsing the Einstein's equation of photoelectric effect,hcλ=KE+ϕKE=1240200−4.2KE=2 eV

Light of wavelength $200 n m$ shines on an aluminium, $4.20 e V$ is required to eject an electron. What is the kinetic energy of the fastest ejected electrons?

The correct option is C $2.00 e V$
Using the Einstein's equation of photoelectric effect,
$\frac{h c}{λ} = K E + ϕ$
$K E = \frac{1240}{200} - 4.2$
$K E = 2 e V$