Question

The energy of a photon of wavelength 2500 is 4.96 eV. When photons of wavelength 3100 incident on a photosensitive material of work function 2.2 eV, the maximum velocity of the emitted photoelectrons is:

• A. $4 \times 10^6 ms^{-1}$
• B. $4 \times 10^5 ms^{-1}$
• C. $8 \times 10^6 ms^{-1}$
• D. $8 \times 10^5 ms^{-1}$

Hint:

Always convert eV to Joules when using mass in kg!

Answer 1

The Correct Option is D

Step 1: List what we know.
A photon of wavelength $2500\ \text{angstrom}$ carries $4.96\ eV$. New photons of wavelength $3100\ \text{angstrom}$ strike a metal of work function $2.2\ eV$. We want the fastest electron's speed.

Step 2: Find the new photon energy.
Photon energy is inversely proportional to wavelength, since $E = \dfrac{hc}{\lambda}$. So $E_{3100} = 4.96 \times \dfrac{2500}{3100} = 4\ eV$.

Step 3: Apply Einstein's equation.
The maximum kinetic energy is the photon energy minus the work function: $K_{max} = E - \Phi = 4 - 2.2 = 1.8\ eV$.

Step 4: Change to joules.
Multiply by $1.6\times10^{-19}$: $K_{max} = 1.8 \times 1.6\times10^{-19} = 2.88\times10^{-19}\ J$.

Step 5: Use the kinetic energy formula.
Since $K_{max} = \dfrac{1}{2}mv^2$, solve for speed: $v = \sqrt{\dfrac{2K_{max}}{m}}$, with electron mass $m = 9\times10^{-31}\ kg$.

Step 6: Compute the speed.
$v = \sqrt{\dfrac{2(2.88\times10^{-19})}{9\times10^{-31}}} = \sqrt{0.64\times10^{12}} = 8\times10^{5}\ ms^{-1}$. \[ \boxed{8 \times 10^5~ms^{-1}} \]