Light of wavelength $4000 \circ A$ is incident on a sodium surface for which the threshold wave length of photo – electrons is $5420 \circ A$ . The work function of sodium is

4.58 eV

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

2.29 eV

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

1.14 eV

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

0.57 eV

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

Open in App

Solution

The correct option is B
2.29 eV

$W_{0} = \frac{12375}{λ_{0} (\circ A)} = \frac{12375}{5420} = 2.28 e V$

Suggest Corrections

Similar questions

Light of wavelength $4000 \circ A$ is incident on a sodium surface for which the threshold wave length of photo – electrons is $5420 \circ A$ . The work function of sodium is

Q. Light of wavelength

4000 0 A

is incident on a sodium surface for which the threshold wavelength of photoelectrons is

5420 0 A

. The work function of sodium is

Q. Light of wavelength 400 angstorm is incident on a surface for which the threshold wavelength photoelectrons is 5420 angstrom. The work function of sodium is ?

Q. U.V. Light of wavelength 300

A^{0}

is incident on sodium surface. Then the maximum velocity of photo-electrons emitted is [neglect the work function of sodium].

If the threshold wavelength for sodium is $5420 \circ A$ , then the work function of sodium is

Join BYJU'S Learning Program

Light of wavelength 4000∘A is incident on a sodium surface for which the threshold wave length of photo – electrons is 5420∘A. The work function of sodium is

The correct option is B 2.29 eV W0=12375λ0(∘A)=123755420=2.28 eV

Light of wavelength $4000 \circ A$ is incident on a sodium surface for which the threshold wave length of photo – electrons is $5420 \circ A$ . The work function of sodium is

The correct option is B
2.29 eV

$W_{0} = \frac{12375}{λ_{0} (\circ A)} = \frac{12375}{5420} = 2.28 e V$