The stopping potential for the photoelectrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Find the wavelength of the radiaiton used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength.

Explanation

To solve the problem step-by-step, we will use the concepts of photoelectric effect and the energy levels of the hydrogen atom.

### Step 1: Understand the given values
- Work function (\( \phi \)) = 1.7 eV
- Stopping potential (\( V_0 \)) = 10.4 V

### Step 2: Use the photoelectric equation
The maximum kinetic energy (\( K_{max} \)) of the emitted photoelectrons can be expressed using the stopping potential:
\[ K_{max} = e \cdot V_0 \]
Where \( e \) is the charge of an electron (approximately 1 eV when using electron volts).

Substituting the values:
\[ K_{max} = 10.4 \, \text{eV} \]

### Step 3: Relate kinetic energy to photon energy
According to Einstein's photoelectric equation:
\[ K_{max} = E_{photon} - \phi \]
Where \( E_{photon} \) is the energy of the incoming photon.

Substituting the known values:
\[ 10.4 \, \text{eV} = E_{photon} - 1.7 \, \text{eV} \]

### Step 4: Solve for \( E_{photon} \)
Rearranging the equation gives:
\[ E_{photon} = 10.4 \, \text{eV} + 1.7 \, \text{eV} = 12.1 \, \text{eV} \]

### Step 5: Calculate the wavelength of the radiation
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E_{photon} = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

Rearranging for \( \lambda \):
\[ \lambda = \frac{hc}{E_{photon}} \]

Substituting the values:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{12.1 \, \text{eV}} \]

Calculating \( \lambda \):
\[ \lambda = \frac{1.241 \times 10^{-6} \, \text{m}}{12.1} \approx 1.022 \times 10^{-7} \, \text{m} = 1022 \, \text{Å} \]

### Step 6: Identify the energy levels in the hydrogen atom
To find the transitions in the hydrogen atom that emit this wavelength, we can use the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where \( R \) is the Rydberg constant (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)).

Using the calculated wavelength:
\[ \frac{1}{1022 \, \text{Å}} = \frac{1}{1022 \times 10^{-10} \, \text{m}} \approx 9.78 \times 10^6 \, \text{m}^{-1} \]

Setting this equal to the Rydberg formula:
\[ 9.78 \times 10^6 = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Solving for \( n_1 \) and \( n_2 \), we find that the transition corresponds to \( n_1 = 1 \) and \( n_2 = 3 \).

### Final Answer
- The wavelength of the radiation used is approximately \( 1022 \, \text{Å} \).
- The energy levels in the hydrogen atom that will emit this wavelength correspond to the transition from \( n = 3 \) to \( n = 1 \).