Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
Advertisement Remove all ads

How Many Wavelengths Are Emitted by Atomic Hydrogen in Visible Range (380 Nm − 780 Nm)? in the Range 50 Nm to 100 Nm? - Physics

Sum

How many wavelengths are emitted by atomic hydrogen in visible range (380 nm − 780 nm)? In the range 50 nm to 100 nm?

Advertisement Remove all ads

Solution

Balmer series contains wavelengths ranging from 364 nm (for n2 = 3) to 655 nm (n_2 = ∞).

So, the given range of wavelength (380−780 nm) lies in the Balmer series.

The wavelength in the Balmer series can be found by

`1/lamda = R (1/2^2 - 1/n^2)`

Here, R =  Rydberg's constant = `1.097 xx 10^7 m^-1`

The wavelength for the transition from n = 3 to n = 2 is given by

`1/lamda_1 = R(1/2^2 - 1/3^2)`

`lamda_1 = 656.3 nm`

The wavelength for the transition from n = 4 to n = 2 is given by

`1/lamda_2 = R(1/2^2 - 1/4^2)`

`lamda_2 =486.1  nm`

The wavelength for the transition from n = 5 to n = 2 is given by

`1/lamda_3 = R (1/2^2 - 1/5^2)`

`lamda_3 = 434.0  nm`

The wavelength for the transition from n = 6 to n = 2 is given by

`1/lamda_4 = R (1/2^2 - 1/6^2)`

`lamda_4 = 410.2  nm`

The wavelength for the transition from n = 7 to n = 2 is given by

`1/lamda_5 = R (1/2^2 - 1/7^2)`

`lamda_5 = 397.0  nm`

Thus, the wavelengths emitted by the atomic hydrogen in visible range (380−780 nm) are 5.

Lyman series contains wavelengths ranging from 91 nm (for n2 = 2) to 121 nm (n_2 = ∞)

So, the wavelengths in the given range (50−100 nm) must lie in the Lyman series.

The wavelength in the Lyman series can be found by

`1/lamda = R(1/1^2 - 1/2^2)`

The wavelength for the transition from n = 2 to n = 1 is given by

`1/lamda_1 = R(1/1^2 - 1/2^2)`

`lamda_1 = 122  nm`

The wavelength for the transition from n = 3 to n = 1 is given by

`1/lamda^2 = R(1/1^2 - 1/2^2)`

`lamda_2 = 103  nm`

The wavelength for the transition from n = 4 to n = 1 is given by

`1/lamda^3= R(1/1^2 - 1/4^2)`

`lamda_3 = 97.3  nm`

The wavelength for the transition from n = 5 to n = 1 is given by

`1/lamda_4 = R(1/1^2 - 1/5^2)`

`lamda_4 = 95.0  nm`

The wavelength for the transition from n = 6 to n = 1 is given by

`1/lamda_5 = R(1/1^2 - 1/6^2)`

`lamda_5 = 93.8  nm`

So, it can be noted that the number of wavelengths lying between 50 nm to 100 nm are 3.

Concept: Introduction of Atoms
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 21 Bohr’s Model and Physics of Atom
Short Answers | Q 1 | Page 382
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×