What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? ( h= Planck's constant, c= velocity of light, R= Rydberg's constant).

Question

What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? ( h= Planck's constant, c= velocity of light, R= Rydberg's constant). (A) (5 texth textc textR/36) (B) (4 texth textc textR/3) (C) (3 texth textc textR/4) (D) (7 texth textc textR/144)

Tardigrade Admin · Accepted Answer

We know that, Δ textE= texth textc⋅ textR[(1/ textn12) - (1/ textn22)] For lowest energy, of the spectral line in Lyman series, textn1=1, textn2=2 Hence, Δ textE= texth textc. textR[(1/12) - (1/22)] Δ textE=(3 texth textc textR/4)

Q. What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series?
( $h =$ Planck's constant, $c =$ velocity of light, $R =$ Rydberg's constant).

$\frac{5 h c R}{36}$

$\frac{4 h c R}{3}$

$\frac{3 h c R}{4}$

$\frac{7 h c R}{144}$

We know that,
$Δ E = h c \cdot R [\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}]$
For lowest energy, of the spectral line in Lyman series, $n_{1} = 1, n_{2} = 2$
Hence,
$Δ E = h c . R [\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}}]$
$Δ E = \frac{3 h c R}{4}$

Q. What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series? ( h= Planck's constant, c= velocity of light, R= Rydberg's constant).

365hcR​

34hcR​

43hcR​

1447hcR​