An atom of hydrogen gas is in an excited stat | vedclass

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(D) The number of spectral lines emitted when an electron transitions from an excited state $n_1$ to the ground state $n=1$ is given by the formula $N

Vedclass · Accepted Answer

$n_1=4, n_2=2$

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(D) The number of spectral lines emitted when an electron transitions from an excited state $n_1$ to the ground state $n=1$ is given by the formula $N = \frac{n_1(n_1-1)}{2}$.Given that the number of emitted wavelengths is $6$,we have $\frac{n_1(n_1-1)}{2} = 6$,which simplifies to $n_1^2 - n_1 - 12 = 0$.Solving this quadratic equation,we get $(n_1-4)(n_1+3) = 0$. Since $n_1$ must be positive,$n_1 = 4$.The electron was initially in state $n_2$ and absorbed a photon to reach $n_1 = 4$. After reaching $n_1$,it returned to the ground state $(n=1)$ emitting $6$ lines. This implies the transition sequence covers all levels from $4$ down to $1$.Since the atom was in an excited state $n_2$ and jumped to $n_1=4$,and the total emission spectrum corresponds to transitions from $n=4$ to $n=1$,the initial state must have been $n_2=2$ to allow for the absorption of a photon to reach $n_1=4$.

An atom of hydrogen gas is in an excited state $n_2$. It absorbs a photon of some energy and jumps to a higher energy state $n_1$. Then,it returns to the ground state after emitting six different wavelengths in the emission spectrum. Determine the values of $n_1$ and $n_2$.

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