Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 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?

#### 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
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#### 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