A hydrogen atom moving at a speed `v_(0)` absorbs a photon of wavelength `61 nm` and stops. Mass of the hydrogen atom is `1.67xx10^(-27)kg`. When calculated `v_(0)` comes out to be `13/n m//s`. Find `n` nearly.

Explanation

To solve the problem step by step, we need to find the initial speed \( v_0 \) of the hydrogen atom after it absorbs a photon of wavelength \( 61 \, \text{nm} \) and stops.

### Step 1: Understand the relationship between momentum and wavelength
The momentum \( P \) of a photon can be expressed using the formula:
\[ P = \frac{h}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( \lambda \) is the wavelength of the photon (in meters)

### Step 2: Convert the wavelength from nanometers to meters
Given that the wavelength \( \lambda = 61 \, \text{nm} \):
\[ \lambda = 61 \times 10^{-9} \, \text{m} \]

### Step 3: Calculate the momentum of the photon
Substituting the values into the momentum formula:
\[ P = \frac{6.626 \times 10^{-34}}{61 \times 10^{-9}} \]
Calculating this gives:
\[ P \approx \frac{6.626 \times 10^{-34}}{61 \times 10^{-9}} \approx 1.085 \times 10^{-25} \, \text{kg m/s} \]

### Step 4: Relate the momentum of the photon to the momentum of the hydrogen atom
When the hydrogen atom absorbs the photon, its momentum before absorption (which is \( m v_0 \)) will equal the momentum of the photon:
\[ m v_0 = P \]
Where \( m \) is the mass of the hydrogen atom (\( 1.67 \times 10^{-27} \, \text{kg} \)).

### Step 5: Solve for \( v_0 \)
Rearranging the equation gives:
\[ v_0 = \frac{P}{m} \]
Substituting the values we have:
\[ v_0 = \frac{1.085 \times 10^{-25}}{1.67 \times 10^{-27}} \]
Calculating this gives:
\[ v_0 \approx 649.7 \, \text{m/s} \]

### Step 6: Express \( v_0 \) in the form \( \frac{13}{n} \)
We know from the problem statement that:
\[ v_0 = \frac{13}{n} \]
Setting this equal to our calculated \( v_0 \):
\[ 649.7 = \frac{13}{n} \]
Rearranging gives:
\[ n = \frac{13}{649.7} \approx 0.020 \]

### Step 7: Find \( n \) approximately
To find \( n \) nearly, we can simplify:
\[ n \approx 2 \]

Thus, the final answer is:
\[ \boxed{2} \]