​For a nuclear spin of spin quantum no.  $$(I=\frac {1}{2})$$ processing in a magnetic field at a Larmor frequency of 300 MHz, the wavelength of incident radiation required to excite the nuclear spins must be approximately:
​

Correct option is C

Explanation-

Given-

​$$\textbf{Larmor frequency } (\nu) = 300\,\text{MHz} = 3 \times 10^8\,\text{Hz}$$

Formula:
The wavelength  and frequency  are related by:​​

$$\lambda = \frac{c}{\nu}$$

Where:

$$c = 3 \times 10^8 \,\text{m/s} \quad \text{(speed of light)} \\\nu = 3 \times 10^8 \,\text{Hz}$$

Substituting values-

                   $$\lambda = \frac{3 \times 10^8 \,\text{m/s}}{3 \times 10^8 \,\text{Hz}} = 1\,\text{m}$$

So, the correct answer is option c : 1 m