If, an electron in hydrogen atom jumps from an orbit of lelvel n=3 to an orbit of level n=2, emitted radiation has a freqwuency (R= Rydbertg's contant ,c = velocity of light)

Options

A. `(3Rc)/(27)`
B. `(3Rc)/(25)`
C. `(8Rc)/(9)`
D. `(5Rc)/(36)`

Explanation

To solve the problem of an electron in a hydrogen atom jumping from an orbit of level \( n = 3 \) to an orbit of level \( n = 2 \) and to find the frequency of the emitted radiation, we can follow these steps:

### Step 1: Identify the Energy Levels
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{R}{n^2} \]
where \( R \) is the Rydberg constant and \( n \) is the principal quantum number.

### Step 2: Calculate the Energy Difference
The energy difference (\( \Delta E \)) between the two levels can be calculated using:
\[ \Delta E = E_2 - E_3 = -\frac{R}{2^2} - \left(-\frac{R}{3^2}\right) \]
Substituting the values:
\[ \Delta E = -\frac{R}{4} + \frac{R}{9} \]

### Step 3: Find a Common Denominator
To combine the fractions, find a common denominator:
\[ \Delta E = \left(-\frac{9R}{36} + \frac{4R}{36}\right) = -\frac{5R}{36} \]
Thus, the energy difference is:
\[ \Delta E = \frac{5R}{36} \]

### Step 4: Relate Energy to Frequency
The energy of the emitted photon is also related to its frequency (\( \nu \)) by the equation:
\[ \Delta E = h \nu \]
where \( h \) is Planck's constant. Therefore, we can express the frequency as:
\[ \nu = \frac{\Delta E}{h} \]

### Step 5: Substitute the Energy Difference
Substituting the value of \( \Delta E \):
\[ \nu = \frac{5R}{36h} \]

### Step 6: Use the Speed of Light Relation
Using the relation between frequency, wavelength, and the speed of light \( c \):
\[ c = \nu \lambda \]
we can express the frequency in terms of the Rydberg constant and the speed of light:
\[ \nu = \frac{5Rc}{36} \]

### Final Answer
Thus, the frequency of the emitted radiation when the electron jumps from \( n = 3 \) to \( n = 2 \) is:
\[ \nu = \frac{5Rc}{36} \]