Question

The first emission line of hydrogen atomic spectrum in the
Balmer series appears is (R = Rydberg constant)
(a) 5/36 R cm⁻¹(b) 3/4 R cm⁻¹
(c) 7/144 R cm⁻¹(d) 9/400 R cm⁻¹

Answer 1

answer : option (1) 5/36 R cm-¹

This question can be solved easily with help of Rydberg's wavelength equation,

formula is... 1/λ = RZ²[1/n1² - 1/n2²]

here R is Rydberg's constant, Z is atomic number, electron transfer from n1 to n2 shells.

for Balmer's series, n1 = 2 and n2 = 3

Z = 1 for hydrogen atom.

now, 1/λ = R(1)² [1/2² - 1/3²]

= R[ 1/4 - 1/9]

= R [(9 - 4)/4 × 9]

= 5R/36 cm-¹

hence option (1) is correct choice.

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Answer 2

The first emission line of hydrogen atomic spectrum in the  Balmer series appears at 5/36 R cm⁻¹

Explanation:

According to Bohr, The wave number can be calculated by the formula

$\bar{v}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \mathrm{cm}^{-1}$

For the first emission line of hydrogen atomic spectrum in the

Balmer series, $n_{1}=2$ and $n_{2}=3$

Therefore,

$\bar{v}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \mathrm{cm}^{-1}$

    $\bar{v}=R\left(\frac{1}{4}-\frac{1}{9}\right) \mathrm{cm}^{-1}$

$\Rightarrow \bar{v} =R\left(\frac{9-4}{4 \times 9}\right) \mathrm{cm}^{-1}$

Therefore, $\bar{v}=\frac{5 R}{36} \mathrm{cm}^{-1}$

Thus, The first emission line of hydrogen atomic spectrum in the  Balmer series appears at 5/36 R cm⁻¹

Hence, the correct answer is option (a)