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Step-by-Step Solution
Step 1: State the photoelectric effect equation
In the photoelectric effect, the maximum kinetic energy of photoelectrons is given by
$K_{\max} = \frac{hc}{\lambda} - \phi$,
where:
$h$ is Planck’s constant.
$c$ is the speed of light.
$\lambda$ is the wavelength of incident light.
$\phi$ is the work function of the material (the minimum energy needed to release an electron).
Since the stopping voltage $V_o$ is related to $K_{\max}$ by
$K_{\max} = e \, V_o$, we can write:
$$
e \, V_o = \frac{hc}{\lambda} - \phi \,.
$$
Step 2: Write the known data for the first wavelength
The first wavelength is $\lambda_1 = \, 670.5 \text{ nm}$, and the corresponding stopping voltage is $V_{o1} = 0.48 \text{ V}$.
Substituting these into the photoelectric equation:
$$
e \, (0.48) = \frac{hc}{670.5 \times 10^{-9}} - \phi \,.
$$
It is common in photoelectric problems to use $hc \approx 1240 \text{ eV·nm}$ when the wavelength is expressed in nanometers and energies in electronvolts. Thus,
$$
e \, (0.48) = \frac{1240 \text{ eV·nm}}{670.5 \text{ nm}} - \phi \,.
$$
We can rearrange this to:
$$
\phi = \frac{1240}{670.5} \text{ eV} - e \, (0.48).
$$
(In practice, you can leave it as is and use it in the subtraction step below.)
Step 3: Write the known data for the second wavelength
The second wavelength is $\lambda_2 = \, 474.6 \text{ nm}$, and its stopping voltage is $V_{o2}$ (unknown). The photoelectric equation becomes:
$$
e \, V_{o2} = \frac{1240 \text{ eV·nm}}{474.6 \text{ nm}} - \phi \,.
$$
Step 4: Subtract the two equations
Subtracting the first equation from the second eliminates $\phi$:
$$
e \,(V_{o2}) - e \,(0.48)
= \frac{1240}{474.6} - \frac{1240}{670.5}.
$$
Factor out $e$ on the left side:
$$
e \,(V_{o2} - 0.48)
= 1240 \left(\frac{1}{474.6} - \frac{1}{670.5}\right) \text{ eV}.
$$
Step 5: Solve for the second stopping voltage
Divide by $e$ (which effectively keeps the values in volts):
$$
V_{o2} - 0.48 = 1240 \left(\frac{1}{474.6} - \frac{1}{670.5}\right).
$$
Numerically evaluating:
1) Compute inside the parentheses:
$$
\frac{1}{474.6} - \frac{1}{670.5}
= \frac{670.5 - 474.6}{474.6 \times 670.5}.
$$
2) Multiply by 1240:
$$
1240 \times \frac{670.5 - 474.6}{474.6 \times 670.5}
\approx 0.76.
$$
3) Therefore,
$$
V_{o2} = 0.48 + 0.76 = 1.24 \,\text{V} \approx 1.25 \,\text{V}.
$$
Step 6: Final Answer
Hence, the stopping voltage for the 474.6 nm wavelength is approximately
$1.25 \,\text{V}$.
