Question

Total number of spectral lines in UV region, during
transition from 5th excited state to 1st excited state
(1) 10
(2) 3
(3) 4
(4) Zero
The first emission line in the atomic spectrum of
hydrogen in the Balmer series appears at -
(1) 5R/36
(2) 3R/4
(3)7R/144 (4)9R/400

pls answer with reason...

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Answer 1

Explanation:

the first question has a direct formula for transition & for 2nd question we have to use formula for balmer transition.

Please let me know for any error.

Answer 2

Answer:

1) The correct answer is option (1).

2)  The correct answer is option (1).

Explanation:

Number of spectral lines formula is given as:

$\frac{(n_2-n_1)(n_2-n_1+1)}{2}$

$n_2$ = Initial level of energy of an electron

$n_1$ = final level of energy of an electron

Given: $n_2= 6$ ( fifth excited level)

$n_1 = 2$ ( first excited level)

$=\frac{(6-2)(6-2+1)}{2}=10$

Total number of spectral lines in UV region, during  transition from 5th excited state to 1st excited state is 10.

2)

Using Rydberg's Equation:

$\bar{\nu}=\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\bar{\nu}$ = Wave number

$\lambda$ = Wavelength of radiatio

$R_H$ = Rydberg's Constant

$n_f$ = final energy level = $2$  

$n_i$= Initial energy level = 3 (Balmer series)

$\bar{\nu}=\frac{1}{\lambda}=R\left(\frac{1}{3^2}-\frac{1}{2^2} \right )$

$\bar{\nu}=\frac{1}{\lambda}=-R\frac{5}{36}$

Negative sign indicates that the energy is released during this transition.

The first emission line in the atomic spectrum of  hydrogen in the Balmer series appears at $R\frac{5}{36}$