Question

An electron in nth orbit of h atom requires 2.856e.v of energy to go to (n+3)th orbit then calculate n

Answer 1

Given

• Electron is in $n^{th}$ orbit.

• 2.856 eV of energy is required to excite to the ${(n+3)}^{th}$

To Find

The value of n.

Concept

The value of energy needed to excite an atom from a lower orbital to a higher orbital is the difference of the energy of each orbital.

Also, Energy of $n^{th}$ orbit of a hydrogen atom( Z = 1) is given by $\frac{ - 13.6}{ {n}^{2} }$.

Calculation

Energy of $n^{th}$ + 2.856 = Energy of ${n +3}^{th}$

=> $\frac{ - 13.6}{ {n}^{2} }$ + 2.856 = $\frac{ - 13.6}{ {(n+3)}^{2} }$.

=> 2.856 = 13.6 ( $\frac{1}{ {n}^{2} } - \frac{1}{ {(n + 3)}^{2} }$ )

=> 2.856 = 13.6 ( $\frac{( {n + 3)}^{2} - {n}^{2} }{ { {n}^{2} + 3n} }$ )

=> 0.21 = ( $\frac{( {n + 3)}^{2} - {n}^{2} }{ { {n}^{2} + 3n} }$ )

=> 0.21 =

$\frac{( {n}^{2} + 6n + 9 - {n}^{2} )}{ { {n}^{2} + 3n} } \\ \\ = > 0.21 = \frac{6n + 9}{ {n}^{2} + 3n } \\ \\ = > 0.7 = \frac{2n + 3}{ {n}^{2} + 3n } \\ \\ = > 7( {n}^{2} + 3n) = 10(2n + 3) \\ \\ = > 7 {n}^{2} + 21n = 20n + 30 \\ \\ = > 7 {n}^{2} + n - 30 = 0 \\ \\ = > 7 {n}^{2} - 14n + 15n - 30 = 0 \\ \\ = > 7n(n - 2) + 15(n - 2) \\ \\ = > (7n + 15)(n - 2) = 0 \\ \\ = > n - 2 = 0 \\ \\ = > n = 2$

$\boxed{ \green{Hence,\:the\:value\:of\:n\:is\:2} }$