Light from a discharge tube containing hydrogen ATOMS falls on a piece of sodium due to the transition of electron from `4^th` orbit to `2^nd` orbit. Work function of sodium is `1.83 eV`. The fastest moving photoelectron is allowed to enter in a magnetic field, which is perpendicular to the direction of motion of photoelectron as shown in the figure. Find the distance (in `mu m` ) covered by the electron in the magnetic field. `B= 1T pi^(2)=10r.`, Mass of electron `=9xx 10^(-31) kg` and `R=` Radius of the path that the most energetic electron takes in the presence of applied magnetic field)
'(##CEN_KSR_PHY_JEE_C29_E01_038_Q01##)'

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the energy of the photon emitted during the electron transition
The energy of the photon emitted when an electron transitions from the 4th orbit to the 2nd orbit in a hydrogen atom is given as \( E = 2.55 \, \text{eV} \).

### Step 2: Calculate the kinetic energy of the ejected photoelectron
The kinetic energy (KE) of the ejected photoelectron can be calculated using the formula:
\[ KE = E_{\text{photon}} - \text{Work Function} \]
Given that the work function of sodium is \( 1.83 \, \text{eV} \):
\[ KE = 2.55 \, \text{eV} - 1.83 \, \text{eV} = 0.72 \, \text{eV} \]

### Step 3: Convert kinetic energy from eV to Joules
To convert the kinetic energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = 0.72 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.152 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the radius of the circular path of the electron in the magnetic field
The radius \( R \) of the circular path of the electron in a magnetic field is given by the formula:
\[ R = \frac{mv}{qB} \]
Where:
- \( m \) is the mass of the electron \( = 9 \times 10^{-31} \, \text{kg} \)
- \( v \) is the velocity of the electron
- \( q \) is the charge of the electron \( = 1.6 \times 10^{-19} \, \text{C} \)
- \( B \) is the magnetic field strength \( = 1 \, \text{T} \)

First, we need to find the velocity \( v \) of the electron using the kinetic energy:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging gives:
\[ v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 1.152 \times 10^{-19} \, \text{J}}{9 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives:
\[ v = \sqrt{\frac{2.304 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{2.56 \times 10^{11}} \approx 1.6 \times 10^6 \, \text{m/s} \]

### Step 5: Substitute values into the radius formula
Now substituting the values into the radius formula:
\[ R = \frac{(9 \times 10^{-31} \, \text{kg}) \times (1.6 \times 10^6 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (1 \, \text{T})} \]
Calculating this gives:
\[ R = \frac{1.44 \times 10^{-24}}{1.6 \times 10^{-19}} = 9 \times 10^{-6} \, \text{m} = 9 \, \mu m \]

### Step 6: Calculate the distance traveled in the magnetic field
The distance traveled by the electron in one complete circular path is given by:
\[ \text{Distance} = 2\pi R \]
Substituting the radius:
\[ \text{Distance} = 2\pi (9 \times 10^{-6} \, \text{m}) \approx 56.52 \times 10^{-6} \, \text{m} = 56.52 \, \mu m \]

### Final Answer
The distance covered by the electron in the magnetic field is approximately \( 56.52 \, \mu m \).

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