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 to be 3.57 V The threshold frequency of the material is

Options

A. `2.5xx10^(15)Hz`
B. `4xx10^(15)Hz`
C. `5xx10^(15)Hz`
D. `1.6xx10^(15)Hz`

Explanation

To solve the problem step by step, we will follow the concepts of energy levels in hydrogen atoms, the photoelectric effect, and the relationship between energy and frequency.

### Step 1: Determine the stopping potential and maximum kinetic energy
The stopping potential \( V \) is given as 3.57 V. The maximum kinetic energy (\( KE_{max} \)) of the photoelectrons can be calculated using the formula:
\[ KE_{max} = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) C). However, since we need the energy in electron volts, we can directly state:
\[ KE_{max} = 3.57 \, \text{eV} \]

**Hint:** The stopping potential directly gives the maximum kinetic energy of the emitted electrons in electron volts.

### Step 2: Calculate the energy levels of the hydrogen atom
The energy of an electron in a hydrogen atom at different energy levels is given by:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]
For the ground state (\( n = 1 \)):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
For the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

**Hint:** Use the formula for energy levels in hydrogen to find the energy at different states.

### Step 3: Calculate the energy released during the transition
The energy released when the electron transitions from the first excited state to the ground state is given by:
\[ E = E_2 - E_1 = (-3.4) - (-13.6) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

**Hint:** The energy released is the difference between the energy levels of the two states.

### Step 4: Use the photoelectric effect equation
According to the photoelectric effect, the energy of the incident photon is related to the maximum kinetic energy of the emitted electrons and the work function (\( \phi \)):
\[ E = KE_{max} + \phi \]
Rearranging gives:
\[ \phi = E - KE_{max} \]
Substituting the values we have:
\[ \phi = 10.2 \, \text{eV} - 3.57 \, \text{eV} = 6.63 \, \text{eV} \]

**Hint:** The work function is the energy required to remove an electron from the material.

### Step 5: Calculate the threshold frequency
The threshold frequency (\( \nu_0 \)) can be calculated using the work function and Planck's constant (\( h \)):
\[ \phi = h \nu_0 \]
Rearranging gives:
\[ \nu_0 = \frac{\phi}{h} \]
Using \( h = 6.63 \times 10^{-34} \, \text{Js} \) and converting \( \phi \) to joules:
\[ \phi = 6.63 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.0608 \times 10^{-18} \, \text{J} \]
Now substituting:
\[ \nu_0 = \frac{1.0608 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} \, \text{Hz} \]

**Hint:** Use the relationship between energy and frequency to find the threshold frequency.

### Final Answer
The threshold frequency of the material is approximately:
\[ \nu_0 = 1.6 \times 10^{15} \, \text{Hz} \]