A standard metre is equal to k wavelengths in vacuum, energy of photon is 2.047eV of the radiation from Krypton 86, where k is

Options

A. 165076.37
B. 16507637.3
C. 1650763.73
D. none of these

Answer

To solve the problem, we need to determine the value of \( k \), which represents the number of wavelengths in a standard meter when the energy of a photon is given as \( 2.047 \, \text{eV} \).

### Step-by-step Solution:

1. **Convert Energy from Electron Volts to Joules**:
The energy of a photon \( E \) is given in electron volts (eV). We need to convert this to joules (J) using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).

\[ E = 2.047 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.278 \times 10^{-19} \, \text{J} \]

2. **Use the Energy-Wavelength Relation**:
The energy of a photon can also be expressed in terms of its wavelength \( \lambda \) using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

3. **Rearranging the Formula to Find Wavelength**:
Rearranging the formula to solve for \( \lambda \):

\[ \lambda = \frac{hc}{E} \]

4. **Substituting Values**:
Now, substitute the values of \( h \), \( c \), and \( E \) into the equation:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3.278 \times 10^{-19} \, \text{J}} \]

\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{3.278 \times 10^{-19} \, \text{J}} \approx 6.06 \times 10^{-7} \, \text{m} = 606 \, \text{nm} \]

5. **Finding the Number of Wavelengths in a Meter**:
To find \( k \), which is the number of wavelengths in one meter, we use the formula:

\[ k = \frac{1 \, \text{m}}{\lambda} \]

Substituting \( \lambda \):

\[ k = \frac{1 \, \text{m}}{6.06 \times 10^{-7} \, \text{m}} \approx 1.65 \times 10^6 \]

6. **Final Answer**:
Thus, the value of \( k \) is approximately \( 1.65 \times 10^6 \).