Question

Light strikes a metal surface, causing photoelectric emission. The stopping potential for the ejected electrons is 7.9 V, and the work function of the metal is 3.3 eV. What is the wavelength of the incident light

Answer 1

Answer:

The wavelength of light is $1.11\times 10^{-7}\ m$.

Explanation:

Given:

Work function of the metal (Ф) = 3.3 eV = $3.3\times 1.6\times 10^{-19}\ J=5.28\times 10^{-19}\ J$

Stopping potential (V) = 7.9 V

Wavelength of light (λ) = ?

We know that for photoelectric effect to take place, the minimum energy of  the incident photon must be equal to the sum of maximum kinetic energy and the work function of the metal surface. This means,

$E_p=K_{max}+\phi$

Now, the kinetic energy of the electrons emitted must be at least equal to the product of stopping potential and the charge on electron of the metal in order to escape the metal surface. Therefore,

$K_{max}=eV$

Where, 'e' is the charge on one electron = $1.6\times 10^{-19}\ C$

Also, the energy of photon is expressed as:

$E_p=\frac{hc}{\lambda}\\\\Where,\\h\to\ Planck's\ constant=6.626\times 10^{-34}\ Js\\c\to\ velocity\ of\ light=3\times 10^{8}\ m/s$

Now, the above equation becomes,

$\frac{hc}{\lambda}=eV+\phi\\\\\lambda=\frac{hc}{eV+\phi}$

Plug in all the given values and solve for λ. This gives,

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{1.6\times10^{-19}\times 7.9+5.28\times 10^{-19}}\\\\\lambda=1.11\times 10^{-7}\ m$

Therefore, the wavelength of light is $1.11\times 10^{-7}\ m$.