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Step-by-Step Solution
Step 1: Understand the Physical Situation
An electron in the hydrogen atom makes a transition from an excited state (quantum number n) to the ground state (n = 1). The photon emitted from this transition illuminates a photosensitive material of work function 2.75 eV, causing photoelectrons to be emitted with a stopping potential of 10 V.
Step 2: Relate Stopping Potential, Work Function, and Photon Energy
The maximum kinetic energy of the photoelectrons, $KE_{\text{max}}$, is given by the stopping potential:
$KE_{\text{max}} = e \times V_{\text{stop}} = 10 \text{ eV},$
where $V_{\text{stop}} = 10 \text{ V}$ and the charge of an electron is $e=1$ (in eV units). The incident photon energy $E_{\gamma}$ is the sum of the work function $\phi$ and the maximum kinetic energy of the emitted electrons:
$E_{\gamma} = \phi + KE_{\text{max}} = 2.75 \text{ eV} + 10 \text{ eV} = 12.75 \text{ eV}.$
Step 3: Identify the Energy Difference in the Hydrogen Atom
The electron in the hydrogen atom initially resides in the energy level $E_{n}$ and drops to the ground state with energy $E_{1}$. The energy of the emitted photon must be
$E_{\gamma} = E_{1} - E_{n},$
(using the sign convention where $E_{n}$ is negative and $E_{1}$ is more negative). For hydrogen, the energy of the level $E_{n}$ is given (in electron volts) by:
$E_{n} = -\frac{13.6 \text{ eV}}{n^{2}}.$
The ground state energy is $E_{1} = -13.6 \text{ eV}.$
Hence, the photon energy emitted in the transition from state n to 1 is:
$E_{\gamma} = \bigl(-\frac{13.6}{1^{2}}\bigr) - \Bigl(-\frac{13.6}{n^{2}}\Bigr) = -13.6 \text{ eV} + \frac{13.6}{n^{2}}.$
Simplifying,
$E_{\gamma} = 13.6 \text{ eV}\Bigl(\frac{1}{n^{2}} - 1\Bigr).$
But we know $E_{\gamma} = 12.75 \text{ eV}$. We check which n satisfies this.
Step 4: Verify the Given Energy Difference for n = 4
For n = 4:
$E_{4} = -\frac{13.6 \text{ eV}}{4^{2}} = -\frac{13.6}{16} \text{ eV} = -0.85 \text{ eV}.$
The ground state energy is $E_{1} = -13.6 \text{ eV}.$
So the energy difference between these levels is
$E_{1} - E_{4} = -13.6 \text{ eV} - (-0.85 \text{ eV}) = -13.6 \text{ eV} + 0.85 \text{ eV} = -12.75 \text{ eV}.$
Since the photon energy is the negative of this difference (because we take the magnitude for emitted photon),
$E_{\gamma} = 12.75 \text{ eV},$
exactly matching the required photon energy for the photoelectric effect. Therefore, n = 4.
Final Answer
The value of n from which the electron jumps to the ground state is 4.
