
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491\text{ }nm$ is $0.71\text{ }V$. When the incident wavelength is changed to a new value, the stopping potential is $1.43\text{ }V$. The new wavelength is:
a. $400\text{ }nm$
b. $382\text{ }nm$
c. $309\text{ }nm$
d. $329\text{ }nm$
Answer
164.1k+ views
Hint: We have to make use of equation of photoelectric emission to make two equations. Then by subtracting them we will be able to find the unknown variable here. The unknown variable is the wavelength here. Planck’s constant $h=6.62\times {{10}^{-34}}\text{ }\dfrac{{{m}^{2}}kg}{s}$ and speed of light $c=3\times {{10}^{8}}\text{ }\dfrac{m}{s}$
Complete answer:
Let us consider that the unknown wavelength be ${{\lambda }_{2}}$ and the known wavelength ${{\lambda }_{1}}=491\text{ }nm$.
The stopping potentials of the electrons are given as $e{{V}_{s1}}=0.71\text{ }V$ and $e{{V}_{s2}}=1.43\text{ }V$
From the equation of photoelectric emission we get,
$\dfrac{hc}{\lambda }=\phi +e{{V}_{s}}$
Where,
${{V}_{s}}$ is the stopping potential and $\phi $ is the work function of the metal.
Hence formulating for the two different stopping potential we get,
$e{{V}_{s1}}=\dfrac{hc}{{{\lambda }_{1}}}-\phi $---(i)
And,
$e{{V}_{s2}}=\dfrac{hc}{{{\lambda }_{2}}}-\phi $---(ii)
Subtracting equation (i) from equation (ii) we get,
$e{{V}_{s1}}-e{{V}_{s2}}=\dfrac{hc}{{{\lambda }_{1}}}-\dfrac{hc}{{{\lambda }_{2}}}$
Arranging the equation we get,
${{V}_{s1}}-{{V}_{s2}}=\dfrac{hc}{e}\left( \dfrac{1}{{{\lambda }_{1}}}-\dfrac{1}{{{\lambda }_{2}}} \right)$
Substituting the values of stopping potentials and wavelength from the given question we get,
$\left( 0.71-1.43 \right)=\dfrac{hc}{e}\left( \dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}} \right)$---(iii)
We know, that the value of $hc=1240\text{ }eV$.
Therefore, substituting the value of $hc$ in $\dfrac{hc}{e}$ we get,
$\dfrac{hc}{e}=\dfrac{1240e}{e}=1240$--(iv)
Substituting the value of equation (iv) in (iii) we get,
$-0.72=1240\left( \dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}} \right)$
Simplifying the equation we get,
$-\dfrac{0.72}{1240}=\dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}}$
Hence again simplifying and calculating it we get,
$\dfrac{1}{{{\lambda }_{2}}}=0.00261$
Inverting it we get,
${{\lambda }_{2}}=383.14\approx 382\text{ }nm$.
So, the correct option is b. $382\text{ }nm$
Note: It must be noted that we have chosen the equation of photoelectric emission as the number of variables are less here. Even we know the value of constant like the Planck’s constant, the speed of light. So, we have easily calculated the value of unknown wavelength.
Complete answer:
Let us consider that the unknown wavelength be ${{\lambda }_{2}}$ and the known wavelength ${{\lambda }_{1}}=491\text{ }nm$.
The stopping potentials of the electrons are given as $e{{V}_{s1}}=0.71\text{ }V$ and $e{{V}_{s2}}=1.43\text{ }V$
From the equation of photoelectric emission we get,
$\dfrac{hc}{\lambda }=\phi +e{{V}_{s}}$
Where,
${{V}_{s}}$ is the stopping potential and $\phi $ is the work function of the metal.
Hence formulating for the two different stopping potential we get,
$e{{V}_{s1}}=\dfrac{hc}{{{\lambda }_{1}}}-\phi $---(i)
And,
$e{{V}_{s2}}=\dfrac{hc}{{{\lambda }_{2}}}-\phi $---(ii)
Subtracting equation (i) from equation (ii) we get,
$e{{V}_{s1}}-e{{V}_{s2}}=\dfrac{hc}{{{\lambda }_{1}}}-\dfrac{hc}{{{\lambda }_{2}}}$
Arranging the equation we get,
${{V}_{s1}}-{{V}_{s2}}=\dfrac{hc}{e}\left( \dfrac{1}{{{\lambda }_{1}}}-\dfrac{1}{{{\lambda }_{2}}} \right)$
Substituting the values of stopping potentials and wavelength from the given question we get,
$\left( 0.71-1.43 \right)=\dfrac{hc}{e}\left( \dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}} \right)$---(iii)
We know, that the value of $hc=1240\text{ }eV$.
Therefore, substituting the value of $hc$ in $\dfrac{hc}{e}$ we get,
$\dfrac{hc}{e}=\dfrac{1240e}{e}=1240$--(iv)
Substituting the value of equation (iv) in (iii) we get,
$-0.72=1240\left( \dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}} \right)$
Simplifying the equation we get,
$-\dfrac{0.72}{1240}=\dfrac{1}{491}-\dfrac{1}{{{\lambda }_{2}}}$
Hence again simplifying and calculating it we get,
$\dfrac{1}{{{\lambda }_{2}}}=0.00261$
Inverting it we get,
${{\lambda }_{2}}=383.14\approx 382\text{ }nm$.
So, the correct option is b. $382\text{ }nm$
Note: It must be noted that we have chosen the equation of photoelectric emission as the number of variables are less here. Even we know the value of constant like the Planck’s constant, the speed of light. So, we have easily calculated the value of unknown wavelength.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
