Question

Light of frequency $5.0 \times 10^{14}$ Hz is incident on a metal surface. If the maximum speed of photoelectrons emitted is $6.63 \times 10^5$ ms$^{-1}$, calculate the threshold frequency for the surface.

Hint:

When dealing with $6.63$ in numerical problems, remember that squaring it gives roughly $44$ ($43.95$). Using standard constants thoughtfully can help avoid rounding errors early in the calculation.

Answer 1

Concept:
• The photoelectric effect is governed by Einstein's photoelectric equation: $K_{max} = h\nu - h\nu_0$.
• Here, $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons, $h\nu$ is the energy of the incident light, and $h\nu_0$ is the work function (minimum energy required to eject an electron).
• $\nu$ is the incident frequency and $\nu_0$ is the threshold frequency.

Step 1: Identify the given parameters
Incident frequency, $\nu = 5.0 \times 10^{14} \text{ Hz}$. Maximum speed of photoelectrons, $v_{max} = 6.63 \times 10^5 \text{ m/s}$. Mass of an electron, $m = 9.1 \times 10^{-31} \text{ kg}$ (standard constant). Planck's constant, $h = 6.63 \times 10^{-34} \text{ J s}$ (standard constant).

Step 2: Calculate the maximum kinetic energy ($K_{max}$)
Using the kinetic energy formula: \[ K_{max} = \frac{1}{2} m v_{max}^2 \]
\[ K_{max} = \frac{1}{2} \times (9.1 \times 10^{-31} \text{ kg}) \times (6.63 \times 10^5 \text{ m/s})^2 \]
\[ K_{max} = 0.5 \times 9.1 \times 10^{-31} \times 43.9569 \times 10^{10} \text{ J} \]
\[ K_{max} = 199.998 \times 10^{-21} \text{ J} \approx 2.0 \times 10^{-19} \text{ J} \]

Step 3: Calculate the energy of the incident photon ($E$)
\[ E = h\nu \]
\[ E = (6.63 \times 10^{-34} \text{ J s}) \times (5.0 \times 10^{14} \text{ Hz}) \]
\[ E = 33.15 \times 10^{-20} \text{ J} = 3.315 \times 10^{-19} \text{ J} \]

Step 4: Calculate the threshold frequency ($\nu_0$)
From Einstein's photoelectric equation: \[ h\nu_0 = h\nu - K_{max} \]
\[ h\nu_0 = 3.315 \times 10^{-19} \text{ J} - 2.0 \times 10^{-19} \text{ J} \]
\[ h\nu_0 = 1.315 \times 10^{-19} \text{ J} \]
Now, solve for $\nu_0$: \[ \nu_0 = \frac{1.315 \times 10^{-19}}{6.63 \times 10^{-34}} \text{ Hz} \]
\[ \nu_0 = 0.19834 \times 10^{15} \text{ Hz} \]
\[ \nu_0 = 1.98 \times 10^{14} \text{ Hz} \]

Step 5: Conclusion
The threshold frequency for the metal surface is approximately $1.98 \times 10^{14} \text{ Hz}$.