Ask & Answer Question

physics ncert exemplar solution class 12th chapter one Photoelectric Emission Dual Nature of Radiation and Matter Physics Class 12th

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

Option 1 -

+3V

Option 2 -

+4V

Option 3 -

-1V

Option 4 -

-3V

4 Views | Posted a week ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a week ago

Correct Option - 4

Detailed Solution:

According to Einstein’s photoelectric equation maximum kinetic energy of photoelectrons, $K E_{max} = E_{v} - φ$

or $2 = 5 - φ$ $∴ φ = 3 e V$

When $E_{v} = 6 e V$ then, $K E_{max} = 6 - 3 = 3 e V$

or $e (V_{c a t h o d e} - V_{a n o d e}) = 3 e V$

or $V_{c a t h o d e} - V_{a n o d e} = 3 V = - V_{s t o p p i n g}$

$∴ V_{s t o p p i n g} = - 3 V$

According to Einstein’s photoelectric equation maximum kinetic energy of photoelectrons, <math> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> max </mrow> </msub> <mo>=</mo> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>v</mi> </mrow> </msub> <mo>−</mo> <mi>φ</mi> </mrow> </math> or   <math> <mrow> <mn>2</mn> <mo>=</mo> <mn>5</mn> <mo>−</mo> <mi>φ</mi> </mrow> </math> <math> <mrow> <mo>∴</mo> <mi>φ</mi> <mo>=</mo> <mn>3</mn> <mi>e</mi> <mi>V</mi> </mrow> </math> When <math> <mrow> <msub> <mrow> <mi>E</mi> </mrow> <mrow> <mi>v</mi> </mrow> </msub> <mo>=</mo> <mn>6</mn> <mi>e</mi> <mi>V</mi> </mrow> </math>  then, <math> <mrow> <mi>K</mi> <msub> <mrow> <mi>E</mi> </mrow> <mrow> max </mrow> </msub> <mo>=</mo> <mn>6</mn> <mo>−</mo> <mn>3</mn> <mo>=</mo> <mn>3</mn> <mi>e</mi> <mi>V</mi> </mrow> </math> or <math> <mrow> <mi>e</mi> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>c</mi> <mi>a</mi> <mi>t</mi> <mi>h</mi> <mi>o</mi> <mi>d</mi> <mi>e</mi> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>a</mi> <mi>n</mi> <mi>o</mi> <mi>d</mi> <mi>e</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>3</mn> <mi>e</mi> <mi>V</mi> </mrow> </math> or   <math> <mrow> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>c</mi> <mi>a</mi> <mi>t</mi> <mi>h</mi> <mi>o</mi> <mi>d</mi> <mi>e</mi> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>a</mi> <mi>n</mi> <mi>o</mi> <mi>d</mi> <mi>e</mi> </mrow> </msub> <mo>=</mo> <mn>3</mn> <mi>V</mi> <mo>=</mo> <mo>−</mo> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>s</mi> <mi>t</mi> <mi>o</mi> <mi>p</mi> <mi>p</mi> <mi>i</mi> <mi>n</mi> <mi>g</mi> </mrow> </msub> </mrow> </math>   <math> <mrow> <mo>∴</mo> <msub> <mrow> <mi>V</mi> </mrow> <mrow> <mi>s</mi> <mi>t</mi> <mi>o</mi> <mi>p</mi> <mi>p</mi> <mi>i</mi> <mi>n</mi> <mi>g</mi> </mrow> </msub> <mo>=</mo> <mo>−</mo> <mn>3</mn> <mi>V</mi> </mrow> </math>

0 Upvotes 0 Downvotes

Similar Questions for you

The work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron whenThe work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron when\

Raj Pandey

For no emission of electron

λ > λ0

In a photoelectric experiment, increasing the intensity of incident light:

Vishal Baghel

Intersity a number of photons kinetic Energy a f

The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are $2.14 e V, 2.30 e V$ and $2.75 e V$ respectively. If incident electromagnetic radiation has an incident energy of $2.20 e V$ , which of these photosensitive surfaces may emit photoelectrons?

Raj Pandey

Incident energy $= 2.20 e V$

If $? < 2.20 e V$ electron will emit.

$? > 2.20 e V$ No electron emission

Only caesium will emit electron

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $^{}$ )

Vishal Baghel

$e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 \times 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 \times 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 \times 10^{- 19}]} \end{array}$

$= 2.27$

When photon of energy $4.0 e V$ strikes the surface of a metal $A$ , the ejected photoelectrons have maximum kinetic energy $T_{A}$ $_{}$ eV and de-Broglie wavelength $λ_{A}$ $_{}$ . The maximum kinetic energy of photoelectrons liberated form another metal B by photon of energy 4.50 $e V$ is $T_{B} = (T_{A} - 1.5) e V$ $_{} (_{})$ . If the de-Broglie wavelength of these photoelectrons $λ_{B} = 2 λ_{A}$ $_{}_{}$ , then the work function of metal $B$ is

Raj Pandey

Relation between De-Broglie wavelength and K.E. is

$λ = \frac{h}{\sqrt[]{2 (K E) m_{e}}} \Rightarrow λ \propto \frac{1}{\sqrt[]{K E}}$

$\begin{array}{r} \frac{λ_{A}}{λ_{B}} = \frac{\sqrt[]{{K E}_{B}}}{\sqrt[]{K E}} \Rightarrow \frac{1}{2} = \sqrt[]{\frac{T_{A} - 1.5}{T_{A}}} \\ \Rightarrow T_{A} 2 e V \\ ∴ {K E}_{B} = 2 - 1.5 = 0.5 e V \\ ?_{B} = 4.5 - 0.5 = 4 e V \end{array}$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question