Question

Calculate the shortest and longest Wavelength in hydrogen spectrum of Lyman series.

Answer 1

AnSwer :-

For Lyman series n₁ = 1

For shortest wavelength in Lyman series .i.e., series limit , the energy difference in two states showing transition should be maximum .i.e., n₂ = ∞

$\sf \dfrac{1}{ \lambda} = R_H \bigg[ \dfrac{1}{ {1}^{2} } - \dfrac{1}{( { \infty )}^{2} } \bigg] = R_H$

$\sf \lambda = \dfrac{1}{3 \times 109678} = 9.117 \times {10}^{ - 6} cm = 911.7 {</p><p>A}^{0}$

For longest wavelength in Lyman series .i.e., first line , the energy difference in two states showing transition should be maximum .i.e., n₂ = 2

$\sf \dfrac{1}{ \lambda } = R_H \bigg[ \dfrac{1}{ {1}^{2} } - \dfrac{1}{ {(2)}^{2} } \bigg ] = \dfrac{3}{4} R_H$

$\sf \lambda = \dfrac{4}{3} \dfrac{1}{R_H } = \dfrac{4}{3 \times 109678}$

$\sf = 1215.7 \times {10}^{ - 8} cm = 1215.7 {A}^{0}$

Answer 2

For Lyman series, n1=1. For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n2=∞. For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n2=2.:

Explanation: