Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured material. The threshold frequency of the material is

Options

A. `2.5 xx 10^(15) Hz`
B. `4 xx 10^(15) Hz`
C. `5 xx 10^(15) Hz`
D. `1.6 xx 10^(15) Hz`

Explanation

To solve the problem step by step, we need to find the threshold frequency of the photosensitive material based on the stopping potential and the energy transitions of the electron in the hydrogen atom.

### Step 1: Determine the energy levels of the hydrogen atom
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

For the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

For the ground state (\( n = 1 \)):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 2: Calculate the energy difference when the electron jumps from the first excited state to the ground state
The energy difference (\( \Delta E \)) when the electron transitions from \( n = 2 \) to \( n = 1 \) is:
\[ \Delta E = E_1 - E_2 = (-13.6 \, \text{eV}) - (-3.4 \, \text{eV}) = -13.6 \, \text{eV} + 3.4 \, \text{eV} = -10.2 \, \text{eV} \]

### Step 3: Relate the energy of the emitted photon to the stopping potential
The energy of the emitted photon when the electron transitions is equal to the energy difference calculated above:
\[ E_{\text{photon}} = \Delta E = 10.2 \, \text{eV} \]

The stopping potential (\( V_s \)) is given as 3.57 V. The maximum kinetic energy (\( KE_{\text{max}} \)) of the emitted photoelectrons is given by:
\[ KE_{\text{max}} = eV_s = 3.57 \, \text{eV} \]

### Step 4: Calculate the threshold frequency
The energy of the photon can also be expressed in terms of its frequency (\( \nu \)) using the equation:
\[ E = h\nu \]
where \( h \) is Planck's constant (\( 4.14 \times 10^{-15} \, \text{eV s} \)).

Setting the energy of the photon equal to the stopping potential energy:
\[ 10.2 \, \text{eV} = h\nu \]

Now, we can solve for the frequency (\( \nu \)):
\[ \nu = \frac{10.2 \, \text{eV}}{h} \]

### Step 5: Calculate the threshold frequency
Using \( h \approx 4.14 \times 10^{-15} \, \text{eV s} \):
\[ \nu = \frac{10.2 \, \text{eV}}{4.14 \times 10^{-15} \, \text{eV s}} \approx 2.46 \times 10^{15} \, \text{Hz} \]

### Final Answer
The threshold frequency of the material is approximately \( 2.46 \times 10^{15} \, \text{Hz} \).

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