The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be

12.4 eV

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

6.2 eV

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

100 eV

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

200 eV

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

Open in App

Solution

The correct option is B
6.2 eV

$K_{m a x} (e V) = 12375 [\frac{1}{λ (\circ A)} - \frac{1}{λ_{0} (\circ A)}]$
$= 12375 [\frac{1}{1000} - \frac{1}{2000}] = 6.2 e V$

Suggest Corrections

Similar questions

Q. The maximum wavelength of radiation that can produce photo-electric effect in a certain metal is

200 n m

. The maximum kinetic energy acquired by electron due to radiation of wavelength

100 n m

will be :

The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be

Q. The maximum wavelength of light for photoelectric effect from a metal is

200 n m

. The maximum kinetic energy of electron which is emitted by the radiation of wavelength

100 n m

will be:

Q. The photoelectric threshold for a certain metal surface is

330 \circ A

. What is the maximum kinetic energy of photoelectron released, if any, by a radiation of wavelength

1100 \circ A

The photosensitive metal is first irradiated with the radiation of wavelength 400 mm and then with radiation of wavelength 800 nm. The change in the maximum kinetic energy of the photoelectron assuming that both wavelengths can cause photoelectric emission, is

Join BYJU'S Learning Program

The maximum wavelength of radiation that can produce photoelectric effect in a certain metal is 200 nm. The maximum kinetic energy acquired by electron due to radiation of wavelength 100 nm will be

The correct option is B 6.2 eV Kmax(eV)=12375[1λ(∘A)−1λ0(∘A)] =12375[11000−12000]=6.2 eV

The correct option is B
6.2 eV

$K_{m a x} (e V) = 12375 [\frac{1}{λ (\circ A)} - \frac{1}{λ_{0} (\circ A)}]$
$= 12375 [\frac{1}{1000} - \frac{1}{2000}] = 6.2 e V$