For intensity I of a light of wavelength 5000 -A the photoelectron saturation current is 0-40 - A and stopping potential is 1-36 V- the work function of metal is

By using E=W_0+K_max E =12375/5000= 2.475 eV and K_max= eV_0=1.36eV So 2.475=W_0 + 1.36 ⇒ W_0=1.1eV ...

112

You visited us 112 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Physics

The Photon Model

For intensity...

Question

For intensity I of a light of wavelength $5000 \circ A$ the photoelectron saturation current is 0.40 $μ A$ and stopping potential is 1.36 V, the work function of metal is

2.47 eV

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

1.36 eV

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

1.10 eV

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

0.43 eV

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

Open in App

Solution

The correct option is C
1.10 eV

$B y u s i n g E = W_{0} + K_{m a x}$
$E = \frac{12375}{5000} = 2.475 e V a n d K_{m a x} = e V_{0} = 1.36 e V$
$S o 2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1 e V$

Suggest Corrections

Similar questions

Q. For intensity

I

of a light of wavelength

5000 ˚ A

the photoelectron saturation current is

0.40 μ A

and stopping potential is

1.36 V

, the work function of metal is :

For intensity I of a light of wavelength $5000 \circ A$ the photoelectron saturation current is 0.40 $μ A$ and stopping potential is 1.36 V, the work function of metal is

Q. A photon of wavelength

100 A^{0}

has energy 12.3 eV. If the wavelength

500 A^{0}

having intensity I falls on a metal surface, the saturation current is

0.40 μ A

and stopping potential is 1.36 V. The work function of the metal is

Q. Given that a photon of light of wavelength

10, 000 o A

has an energy equal to 1.23 eV. When light of wavelength

5000 o A

and intensity

I_{0}

falls on a photoelectric cell, the saturation current is

0.40 \times 10^{- 6} A

and the stopping potential is 1.36 V; then the work function is :

A monochromatic light source of intensity 5 mW emits $8 \times 10^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Join BYJU'S Learning Program

For intensity I of a light of wavelength 5000∘A the photoelectron saturation current is 0.40 μ A and stopping potential is 1.36 V, the work function of metal is

The correct option is C 1.10 eV By using E=W0+Kmax E=123755000=2.475 eV and Kmax=eV0=1.36eV So 2.475=W0+1.36⇒W0=1.1eV

For intensity I of a light of wavelength $5000 \circ A$ the photoelectron saturation current is 0.40 $μ A$ and stopping potential is 1.36 V, the work function of metal is

The correct option is C
1.10 eV

$B y u s i n g E = W_{0} + K_{m a x}$
$E = \frac{12375}{5000} = 2.475 e V a n d K_{m a x} = e V_{0} = 1.36 e V$
$S o 2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1 e V$