Question

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 1

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.