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12.Atoms
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In a hydrogen atom, the electron is in $n^{th}$ excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of $n$ :
A
$6$
B
$7$
C
$8$
D
$5$
Solution
Number of possible emission lines are $n(n-1) / 2$ when an electron jumps from $n^{t h}$ state to ground state. In this question, this value should be
$(n-1)(n-2) / 2$
Hence, $10=(n-1) \frac{n-2}{2}$
Solving this, we getn $=6$
Standard 12
Physics
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