Question

Which of the following should be the correct value of the wave number of first line in Balmer series of hydrogen atom? (a) 5R/36 (b) –5R/36 (c) R/9 (d) –R/9

Answer 1

Answer:

Balmer series starts from ground level as n=2

So, it's first line will be of the electron transition from n=2 to n=3

Using Rydberg's formula,

wavenumber = R(1/2^2-1/3^2 ) = R(1/4-1/9) = 5R/36 So, it'll be 5R/36.

Hope this helps you

Answer 2

The value of the wave number of the first line of the Balmer series is 5R/36, and option 'A' is correct.

Given:-

Atom = hydrogen

Line of Balmer series = 1st line

To Find:-

The value of the wave number of the first line of the Balmer series.

Solution:-

We can easily find out the value of the wave number of the first line of the Balmer series by using these simple steps.

As

Atom = hydrogen

Line of Balmer series = 1st line

Since it's a Balmer series, the value of a lower orbital must be 2.

i.e. n1 = 2

and for the first line, n2 = 3

z for H2 = 1

Wave number (v) =?

Now, according to the formula for wave number

$v =R \times z( \frac{1}{ {(n1)}^{2} } - \frac{1}{ {(n2)}^{2} } )$

$v =R \times 1 \times ( \frac{1}{ {(2)}^{2} } - \frac{1}{ {(3)}^{2} } )$

$v =R \times ( \frac{1}{ 4 } - \frac{1}{ 9 } )$

$v = R \times ( \frac{9 - 4}{4 \times 9} )$

$v = R \frac{5}{36}$

$v = \frac{5R}{36}$

Hence, The value of the wave number of the first line of the Balmer series is 5R/36, and option 'A' is correct.

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