QJEEM23Ph3429: JEE Main 2023 Physics - Class 12 JEEM23Ph3429 Solution: A monochromatic light is incident on a hydrogen sample in…

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

$$E_n=-\frac{13.6\mathrm{eV}}{n^2}$$

where $$E_n$$ is the energy of the nth level and $$n$$ is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

1. From n = 2 to n = 1
2. From n = 3 to n = 1
3. From n = 3 to n = 2
4. From n = 4 to n = 1
5. From n = 4 to n = 2
6. From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

$$\mathrm{\Delta }E=E_4-E_1=\frac{13.6\mathrm{eV}}{4^2}-\frac{13.6\mathrm{eV}}{1^2}=-0.85\mathrm{eV}+13.6\mathrm{eV}=12.75\mathrm{eV}$$

The frequency of the incident light is related to the energy difference by the equation:

$$\mathrm{\Delta }E=h\nu$$

where $$h$$ is the Planck's constant and $$\nu$$ is the frequency.

Now, we can solve for the frequency:

$$\nu =\frac{\mathrm{\Delta }E}{h}=\frac{12.75\mathrm{eV}}{4.25\times {10}^{-15}\mathrm{eVs}}=3\times {10}^{15}\mathrm{Hz}$$

So, the value of $$x$$ is 3.