Light of wavelength 4000 A ∘ and intensity 100 Wm - 2 is incident on a plate of threshold frequency 5 . 5 × 10 14 Hz . Find the maximum kinetic energy of photoelectrons and number of photons incident per m 2 per second.

K . E = h v - h v ∘ K . E = h c l - h ( 5 . 5 × 10 14 ) K . E = 6 . 626 × 10 - 34 3 × 10 8 4000 × 10 - 10 - 5 . 5 × 10 14 J K . E = 1 . 32 × 10 - 19 J Energy = 100 J ⇒ nhc l = 100 ⇒ n × 6 . 626 × 10 - 34 × 3 × 10 8 4000 × 10 - 10 = 100 ⇒ n = 2 . 02 × 10 20 m 2

Light of wavelength 4000 A ∘ and intensity 100 Wm - 2 is incident on a plate of threshold frequency 5 . 5 × 10 14 Hz . Find the maximum kinetic energy of photoelectrons and number of photons incident per m 2 per second.

K . E = h v - h v ∘ K . E = h c l - h ( 5 . 5 × 10 14 ) K . E = 6 . 626 × 10 - 34 3 × 10 8 4000 × 10 - 10 - 5 . 5 × 10 14 J K . E = 1 . 32 × 10 - 19 J Energy = 100 J ⇒ nhc l = 100 ⇒ n × 6 . 626 × 10 - 34 × 3 × 10 8 4000 × 10 - 10 = 100 ⇒ n = 2 . 02 × 10 20 m 2

The potential difference between the cathode and the target electrode in a Coolidge tube is changed from 20 kV to 30 kV . By what percent the minimum wavelength of the emitted X-rays will change?

When electrons accelerated through a potential difference V Strike a target, the maximum frequency of the emitted X-rays is given by eV = hv max Where, e Is charge of the electron and h The Planck constant. But v max = c λ min ∴ eV = hc λ min OR λ min = hc eV And hc e = k Constant Then, λ min = k v ∴ λ min 1 = k 20 . . . . . . . . . . . . ( i ) And λ min 2 = k 30 . . . . . . . . . . . . . . . . . . . ( ii ) So, the change is ∆ λ min = λ min 2 - λ min 1 ∆ λ min = k 30 - k 20 = - k 60 Then for percent the minimum wavelength of the emitted X-rays will change ∆ λ min λ min 1 × 100 = - k 60 k 20 × 100 = - 20 60 × 100 = - 33 . 3 % = 33 . 3 %

MEDIUM

12th West Bengal Board

IMPORTANT

Earn 100

Light of wavelength $4000 \overset{\circ}{A}$ and intensity $100 {Wm}^{- 2}$ is incident on a plate of threshold frequency $5.5 \times 10^{14} Hz$ . Find the maximum kinetic energy of photoelectrons and number of photons incident per $m^{2}$ per second.

Important Questions on Photoelectric Emission and Production and Properties of X-Rays

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 16

What should be the minimum potential difference between the ends of a coolidge tube, to produce X-rays of wavelength

0.8 \vec{A}

?
Given:

h = 6.62 \times 10^{- 34} Js, e = 1.60 \times 10^{- 19}

C and

c = 3 \times 10^{8} m / s

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 17

Calculate the minimum wavelength of X-rays produced by an X -ray tube operating at

30 kV . [h = 6.6 \times 10^{- 27} erg . c]

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 18

The potential difference across an X -ray tube is

12.4 kV

and the current through the filament is

2 mA .

Calculate the number of electrons striking per second on the anode

HARD

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 18

The potential difference across an X -ray tube is

12.4 kV

and the current through the filament is

2 mA .

Calculate the speed with which the electron strikes the target

EASY

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 18

The potential difference across an X -ray tube is

12.4 kV

and the current through the filament is

2 mA .

Calculate the shortest wavelength of X -rays emitted.

HARD

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 19

The potential difference between the cathode and the target electrode in a Coolidge tube is changed from

20 kV

30 kV

. By what percent the minimum wavelength of the emitted X-rays will change?

HARD

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 20

When a beam of

10.6 eV

photons of intensity

2.0 W / m^{2}

falls on a platinum surface of area

1.0 \times 10^{- 4} m^{2}

and work function

5.6 eV, 0.53 %

of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in

eV

). Take

1 eV = 1.6 \times 10^{- 19} J

MEDIUM

12th West Bengal Board

IMPORTANT

A Text Book of PHYSICS PART 2 : CLASS 12>Chapter 19 - Photoelectric Emission and Production and Properties of X-Rays>EXERCISE 19>Q 21

A silver sphere of radius

1 cm

and work function

4.7 eV

is suspended from an insulating thread in free space. It is under continuous illumination of

200 nm

wavelength light. As photoelectrons are emitted, the sphere get charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is

A \times 10^{Z}

(where

1 ⟨ A < 10

). Find the value of Z.

Light of wavelength 4000 A∘ and intensity 100Wm-2 is incident on a plate of threshold frequency 5.5×1014 Hz. Find the maximum kinetic energy of photoelectrons and number of photons incident per m2 per second.

Important Questions on Photoelectric Emission and Production and Properties of X-Rays

Learn and Prepare for any exam you want !

Light of wavelength $4000 \overset{\circ}{A}$ and intensity $100 {Wm}^{- 2}$ is incident on a plate of threshold frequency $5.5 \times 10^{14} Hz$ . Find the maximum kinetic energy of photoelectrons and number of photons incident per $m^{2}$ per second.