The wave number of the radiation whose quantum is 1 erg is

Options

A. `5 xx 10^(15) cm^(-1)`
B. `15 xx 10^(5) cm^(-1)`
C. `1.5 xx 10^(15) cm^(-1)`
D. `5 xx 10^(5) cm^(-1)`

Answer

To solve the question regarding the wave number of radiation whose quantum is 1 erg, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Energy from Ergs to Joules:**
- Given energy \( E = 1 \) erg.
- We know that \( 1 \) joule \( = 10^7 \) ergs.
- Therefore, \( 1 \) erg \( = 10^{-7} \) joules.
- Thus, \( E = 1 \text{ erg} = 10^{-7} \text{ joules} \).

2. **Use the Formula for Energy:**
- The energy of a photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
- Where:
- \( E \) is the energy,
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \),
- \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \),
- \( \lambda \) is the wavelength.

3. **Rearranging the Formula:**
- We need to find the wave number \( \bar{\nu} \), which is defined as:
\[ \bar{\nu} = \frac{1}{\lambda} \]
- Rearranging the energy formula gives us:
\[ \frac{E}{hc} = \frac{1}{\lambda} = \bar{\nu} \]
- Thus, we can express wave number as:
\[ \bar{\nu} = \frac{E}{hc} \]

4. **Substituting Values:**
- Substitute \( E = 10^{-7} \) joules, \( h = 6.626 \times 10^{-34} \) J s, and \( c = 3 \times 10^8 \) m/s into the equation:
\[ \bar{\nu} = \frac{10^{-7}}{(6.626 \times 10^{-34})(3 \times 10^8)} \]

5. **Calculating the Denominator:**
- Calculate \( hc \):
\[ hc = (6.626 \times 10^{-34}) \times (3 \times 10^8) = 1.9878 \times 10^{-25} \text{ J m} \]

6. **Calculating Wave Number:**
- Now substitute back to find \( \bar{\nu} \):
\[ \bar{\nu} = \frac{10^{-7}}{1.9878 \times 10^{-25}} \approx 5.03 \times 10^{15} \text{ m}^{-1} \]
- Convert \( \bar{\nu} \) to cm\(^{-1}\) (since \( 1 \text{ m} = 100 \text{ cm} \)):
\[ \bar{\nu} = 5.03 \times 10^{15} \text{ m}^{-1} \times \frac{1 \text{ m}}{100 \text{ cm}} = 5.03 \times 10^{13} \text{ cm}^{-1} \]

7. **Final Answer:**
- Rounding gives us \( \bar{\nu} \approx 5 \times 10^{15} \text{ cm}^{-1} \).

### Conclusion:
The wave number of the radiation whose quantum is 1 erg is approximately \( 5 \times 10^{15} \text{ cm}^{-1} \).