JEE MAIN - Physics (2023 - 12th April Morning Shift - No. 15)
A $$12.5 \mathrm{~eV}$$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:
2
4
3
1
Explanation
The energy of an electron in an excited state of hydrogen is given by the equation:
Code snippet
$$E = -13.6 \frac{1}{n^2} \mathrm{~eV}$$
where n is the principal quantum number of the state.
The energy of the electron beam is 12.5 eV, which is enough to excite the electron to the n = 3 state.
The possible transitions from n = 3 to lower energy states are:
n = 3 to n = 2, with a wavelength of 656.33 nm (H-alpha)
n = 3 to n = 1, with a wavelength of 102.57 nm (Lyman-alpha)
n = 2 to n = 1, with a wavelength of 121.57 nm (Lyman-beta)
Therefore, there are 3 possible spectral lines that can be emitted.
Code snippet
$$E = -13.6 \frac{1}{n^2} \mathrm{~eV}$$
where n is the principal quantum number of the state.
The energy of the electron beam is 12.5 eV, which is enough to excite the electron to the n = 3 state.
The possible transitions from n = 3 to lower energy states are:
n = 3 to n = 2, with a wavelength of 656.33 nm (H-alpha)
n = 3 to n = 1, with a wavelength of 102.57 nm (Lyman-alpha)
n = 2 to n = 1, with a wavelength of 121.57 nm (Lyman-beta)
Therefore, there are 3 possible spectral lines that can be emitted.
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