Photoemission from a metal surface takes place for frequencies $\nu_1$ and $\nu_2$ of incident rays ($\nu_1 > \nu_2$). The maximum kinetic energy of the photoelectrons emitted is in the ratio $1 : K$. The threshold frequency ($\nu_0$) of the metallic surface is
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To verify your algebraic steps quickly during an exam, check the dimensions or units of your final expression. The denominator $(K - 1)$ is a dimensionless scalar ratio, while the numerator $(K\nu_1 - \nu_2)$ carries units of frequency ($\text{s}^{-1}$). This confirms the final expression has the correct physical units for frequency!
Step 1: The setting.
Light of two frequencies $\nu_1$ and $\nu_2$ (with $\nu_1 > \nu_2$) hits the same metal and knocks out electrons. The largest kinetic energies in the two cases are in the ratio $1 : K$. We must find the threshold frequency $\nu_0$ of the metal.
Step 2: The photoelectric equation.
Einstein's rule says the maximum kinetic energy is the photon energy minus the energy needed to escape:
\[ K.E. = h(\nu - \nu_0) \]
Step 3: Write both cases.
\[ (K.E.)_1 = h(\nu_1 - \nu_0), \qquad (K.E.)_2 = h(\nu_2 - \nu_0) \]
Step 4: Use the given ratio.
The ratio is $1 : K$, so
\[ \frac{(K.E.)_1}{(K.E.)_2} = \frac{1}{K} \;\Rightarrow\; \frac{\nu_1 - \nu_0}{\nu_2 - \nu_0} = \frac{1}{K} \]
The $h$ cancels.
Step 5: Cross multiply and collect.
\[ K(\nu_1 - \nu_0) = \nu_2 - \nu_0 \]
\[ K\nu_1 - K\nu_0 = \nu_2 - \nu_0 \]
Bring the $\nu_0$ terms together:
\[ K\nu_1 - \nu_2 = K\nu_0 - \nu_0 = \nu_0(K - 1) \]
Step 6: Solve for the threshold.
\[ \nu_0 = \frac{K\nu_1 - \nu_2}{K - 1} \]
This is option (4).
\[ \boxed{\nu_0 = \frac{K\nu_1 - \nu_2}{K - 1}} \]