The value of Planck's constant is $6.63 \times 10^{-34} Js$. The speed of light is $3 \times 10^{17} nm \, s^{-1}$. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of $6 \times 10^{15} \, s^{-1}$ ?

Question

The energy of the nth orbit of the hydrogen atom is given by: \[ E_n = - \frac{13.6}{n^2} \, \text{eV} \] What is the energy of the second orbit (n = 2) of the hydrogen atom?

Answer 1

To find the wavelength of a quantum of light with a given frequency, we use the relationship between the speed of light, wavelength, and frequency, expressed by the formula:

c = \lambda \cdot \nu

Where:

c = speed of light
\lambda = wavelength of light
\nu = frequency of light

Given:

Speed of light, c = 3 \times 10^{17} \, \text{nm s}^{-1}
Frequency, \nu = 6 \times 10^{15} \, \text{s}^{-1}

We need to solve for the wavelength \lambda:

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

Substitute the given values into the equation:

\lambda = \frac{3 \times 10^{17} \, \text{nm s}^{-1}}{6 \times 10^{15} \, \text{s}^{-1}}

Calculate \lambda:

\lambda = 0.5 \times 10^{2} \, \text{nm}

\lambda = 50 \, \text{nm}

Thus, the closest value to the wavelength of the quantum of light is 50 nm, which matches the correct answer.

Therefore, the correct option is 50.

Answer 2