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Step-by-Step Detailed Solution
Step 1: Identify the Relevant Photoelectric Effect Formula
The maximum kinetic energy $K_{ \max }$ of an ejected photoelectron can be written as:
$K_{\max} = \dfrac{hc}{\lambda} - \phi,
where $h$ is Planck’s constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident photon, and $\phi$ is the work function of the cathode material. Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{\max} = e \, V_s$, we can write:
$e \, V_s = \dfrac{hc}{\lambda} - \phi.
Step 2: Convert Constants into Convenient Units
In electronvolts (eV), a commonly used approximation is
$ \dfrac{hc}{1 \text{ nm}} \approx 1240 \text{ eV·nm}$.
We also have the work function $\phi = 2.5 \text{ eV}$.
Step 3: Calculate Stopping Potential for $\lambda = 280\text{ nm}$
Substitute $\lambda = 280\text{ nm}$ into
$e \, V_{s_1} = \dfrac{1240 \text{ eV·nm}}{280 \text{ nm}} - 2.5 \text{ eV}$:
$e \, V_{s_1} = \dfrac{1240}{280} - 2.5 = 4.43 - 2.5 \approx 1.93 \text{ eV}.
Thus,
$V_{s_1} = 1.93 \text{ V}.
Step 4: Calculate Stopping Potential for $\lambda = 400\text{ nm}$
Substitute $\lambda = 400\text{ nm}$ into
$e \, V_{s_2} = \dfrac{1240 \text{ eV·nm}}{400 \text{ nm}} - 2.5 \text{ eV}$:
$e \, V_{s_2} = \dfrac{1240}{400} - 2.5 = 3.10 - 2.5 = 0.60 \text{ eV}.
Thus,
$V_{s_2} = 0.60 \text{ V}.
Step 5: Find the Change in Stopping Potential
The change in stopping potential $\Delta V$ is the difference between $V_{s_1}$ and $V_{s_2}$:
$\Delta V = V_{s_1} - V_{s_2} = 1.93 \text{ V} - 0.60 \text{ V} = 1.33 \text{ V}.
Rounding suitably, we often express this as approximately $1.3 \text{ V}.$
Step 6: Conclude the Answer
The change in stopping potential when the incident radiation is switched from 280 nm to 400 nm is about
1.3 V.
