Question

The work function of a metal is $1\, eV$. Light of wavelength $3000 \,?$ is incident on this metal surface. The velocity of emitted photoelectrons will be

• A. $10 \,ms ^{-1}$
• B. $1 \times 10^{3} ms ^{-1}$
• C. $1 \times 10^{4} ms ^{-1}$
• D. $1 \times 10^{6} ms ^{-1}$

Answer 1

The Correct Option is D

To determine the velocity of the emitted photoelectrons, we can apply the photoelectric effect formula:

\(E = h \nu = \frac{hc}{\lambda}\) 

Here, \(E\) is the energy of the incident photons, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, \(c\) is the speed of light, and \(\lambda\) is the wavelength.

Given:

• Work function, \(\phi = 1\, eV\)
• Wavelength, \(\lambda = 3000 \, \mathring{A} = 3000 \times 10^{-10} m\)
• The energy equivalent of \(1\, eV = 1.6 \times 10^{-19} J\)

First, let's convert the work function from eV to joules:

\(\phi = 1\, eV = 1.6 \times 10^{-19} J\)

Using Planck's constant, \(h = 6.626 \times 10^{-34} Js\), and the speed of light, \(c = 3 \times 10^{8} m/s\), we calculate the energy of the incident light:

\(E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{3000 \times 10^{-10}}\)

Calculating the above expression:

\(E = \frac{19.878 \times 10^{-26}}{3000 \times 10^{-10}} = 6.626 \times 10^{-19} J\)

This is the energy of the incident photon. The kinetic energy of the emitted photoelectron is calculated using the equation:

\(\frac{1}{2}mv^{2} = E - \phi\)

Substitute the known values:

\(\frac{1}{2}mv^{2} = 6.626 \times 10^{-19} - 1.6 \times 10^{-19} = 5.026 \times 10^{-19} J\)

Now solve for the velocity, \(v\), using the electron's mass \(m = 9.1 \times 10^{-31} kg\):

\(v = \sqrt{\frac{2 \times 5.026 \times 10^{-19}}{9.1 \times 10^{-31}}}\)

Calculating the above expression:

\(v = \sqrt{\frac{10.052 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{1.104 \times 10^{12}} \approx 1.05 \times 10^{6} ms^{-1}\)

Thus, the velocity of the emitted photoelectrons is approximately \(1 \times 10^{6} ms^{-1}\). Hence, the correct option is:

• \(1 \times 10^{6} ms^{-1}\)

Answer 2

$E _{ k \max } =\left( E _{ photon }- W \right) eV$
$=\left(\frac{12.42 \times 10^{-7}}{3 \times 10^{-7}}-1\right) eV$
$=(4-1) eV $
$E _{ k \max } =3 \,eV =3 \times 1.6 \times 10^{-19} J $
$ \frac{1}{2} mv _{\max }^{2} =3 \times 1.6 \times 10^{-19} $
$ v _{\max }^{2} =\frac{3 \times 2 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$
$v _{\max }^{2} =10^{12} $
$ v _{\max } =10^{6} m s ^{-1} $