Question

Electromagnetic radiations having wavelength 310 Å are subjected to a metal sheet having work function of 12.8 eV. What will be the velocity of photoelectrons with maximum kinetic energy.

• A. 0, no emission will occur
• B. 2.18 x $10^6$m/s
• C. 2.18$\sqrt{2}$ x $10^6$m/s
• D. 8.72 x $10^6$m/s

Answer 1

Answer: (C)

2.18$\sqrt{2}$ x $10^6$m/s

Answer 2

The problem asks for the maximum velocity of photoelectrons ejected from a metal sheet when electromagnetic radiation of a given wavelength is incident on it. We are provided with the wavelength of the incident radiation and the work function of the metal.

1. Calculate the energy of the incident photons ($E$):

The wavelength of the incident radiation is $\lambda = 310 \text{ Å}$.
We can convert this to meters: $\lambda = 310 \times 10^{-10} \text{ m}$.
The energy of a photon can be calculated using the formula $E = \frac{hc}{\lambda}$.
It's often convenient to use the constant $hc = 12400 \text{ eV Å}$.
$E = \frac{12400 \text{ eV Å}}{310 \text{ Å}} = 40 \text{ eV}$.

2. Check for photoelectron emission:

The work function of the metal is $W = 12.8 \text{ eV}$.
Since the energy of the incident photon ($E = 40 \text{ eV}$) is greater than the work function ($W = 12.8 \text{ eV}$), photoemission will occur.

3. Calculate the maximum kinetic energy of the photoelectrons ($KE_{max}$):

According to Einstein's photoelectric equation:
$KE_{max} = E - W$
$KE_{max} = 40 \text{ eV} - 12.8 \text{ eV} = 27.2 \text{ eV}$.

4. Convert maximum kinetic energy to Joules:

To use the kinetic energy formula $KE = \frac{1}{2}mv^2$, we need $KE_{max}$ in Joules. Use the conversion factor $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$.
$KE_{max} = 27.2 \text{ eV} \times (1.602 \times 10^{-19} \text{ J/eV})$
$KE_{max} = 43.5744 \times 10^{-19} \text{ J} = 4.35744 \times 10^{-18} \text{ J}$.

5. Calculate the velocity of the photoelectrons ($v$):

The mass of an electron is $m_e = 9.109 \times 10^{-31} \text{ kg}$.
Using the kinetic energy formula $KE_{max} = \frac{1}{2}m_ev^2$:
$v^2 = \frac{2 \times KE_{max}}{m_e}$
$v^2 = \frac{2 \times 4.35744 \times 10^{-18} \text{ J}}{9.109 \times 10^{-31} \text{ kg}}$
$v^2 = \frac{8.71488 \times 10^{-18}}{9.109 \times 10^{-31}}$
$v^2 = 0.956735 \times 10^{13} \text{ m}^2/\text{s}^2$
$v^2 = 9.56735 \times 10^{12} \text{ m}^2/\text{s}^2$

Now, take the square root to find $v$:
$v = \sqrt{9.56735 \times 10^{12}} \text{ m/s}$
$v \approx 3.0931 \times 10^6 \text{ m/s}$.

Our calculated value $3.0931 \times 10^6 \text{ m/s}$ is closest to $3.0829 \times 10^6 \text{ m/s}$. The small difference is due to rounding of constants and values in the options.