Question

The stopping potential for electrons emitted from a photosensitive surface illuminated by the light of wavelength $491nm$ is $0.710V$. When the incident wavelength is changed to a new value, the stopping potential is $1.43V$. What is this new wavelength?

• A. $400nm$
• B. $382nm$
• C. $309nm$
• D. $329nm$

Answer 1

The correct option is B

$382nm$

Step 1: Given Data

The first wavelength, ${\lambda }_1=491nm$

The first stopping potential $V_1=0.710V$

The second stopping potential $V_2=1.43V$

Let ${\lambda }_2$ be the second wavelength.

Let the maximum kinetic energy be $K_m$.

Let $\varphi$ be the work function of the material.

Let $c$ be the speed of light.

Let $h$ be Planck's constant.

Step 2: Formula Used

We use the photoelectric effect equation in the form

$\frac{hc}{\lambda }=\varphi +K_m$.

The work function depends only on the material and the condition of the surface, and not on the wavelength of the incident light.

Also, the maximum kinetic energy of the electrons equals the stopping voltage when measured in electron volt.

Step 3: Calculate the new Wavelength

Assume $K_{m_1}=0.710eV$ be the maximum kinetic energy of electrons ejected by light with the first wavelength, and

${K_m}_{{}_2}=1.43eV$ be the maximum kinetic energy of electrons ejected by light with the second wavelength. Then,

$\frac{hc}{{\lambda }_1}=\varphi +K_{m1},\frac{hc}{{\lambda }_2}=\varphi +K_{m2}$

The first equation yields $\varphi =\frac{hc}{{\lambda }_1}-K_{m_1}$. When this is used to substitute for $\varphi$ in the second equation, the result is

$\left(\frac{hc}{{\lambda }_2}\right)=\left(\frac{hc}{{\lambda }_1}\right)-K_{m1}+K_{m2}$

We know that $hc=1240eVnm$

The solution for ${\lambda }_2$ is

${\lambda }_2=\frac{hc{\lambda }_1}{hc+{\lambda }_1\left(K_{m2}-K_{m1}\right)}=\frac{\left(1240V.nm\right)\left(491nm\right)}{1240eV.nm+\left(491nm\right)\left(1.43eV-0.710eV\right)}=382nm$

Hence, the correct answer is option (B).