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Line Spectra

In an atom an...

Question

In an atom an electron excites to the fourth orbit.When it jumps back to the energy level one a spectrum is formed. Total number of spectral lines in this spectrum would be

A

3

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B

4

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C

5

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D

6

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Solution

The correct option is D 6
The relation to find number of spectral lines is $\frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2}$

Here transition is from n = 4 to n = 1

So, number of spectral lines = $\frac{3 \times 4}{2} = 6$

Hence, there are 6 spectral lines.

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Q. In atom an electron excites to the fourth orbit. When it jumps back to the energy levels a spectrum is formed. Total number of spectral lies in this spectrum would be:

Q. The number of spectral line obtained in Bohr spectrum of hydrogen atom when an electron is excited from ground level to 5th orbit is :

Q. When an electron in hydrogen spectrum jumps from

n = 7

n = 2

, the total number of spectral lines possible are:

Q. When the electron jumps from

5^{t h}

orbit to ground state, the number of spectral lines produced in the hydrogen spectrum is:

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. The maximum number of lines produced when an electron jumps from nth level to ground level is equal Maximum number of lines produced when an electron jumps from nth level to ground level is equal to.

\frac{1 (n - 1)}{2} .

For example, in the case of

n = 4,

number of lines produces is 6.

(4 \to 3, 4 \to 2, 4 \to 1, 3 \to 2, 3 \to 1, 2 \to 1) .

When an electron returns from to state, the number of lines in the spectrum will be equal to

\frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2}

If the electron comes back from energy level having energy to energy level having energy, then the difference may be expressed in the terms of energy of photon as

E_{2} - E_{1} Δ E, λ = \frac{h c}{Δ E}

Since h and c are constants,

Δ E

corresponds to definite energy; thus each transition from one energy level to another will produce a light of definite wavelength. This is observed as a line in the spectrum of the hydrogen atom. Wavenumber of line is given by the formula

¯ v = (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) .

where R is a Rydberg's constant

(R = 1.1 \times 10^{7} m^{- 1})

The energy photon emitted corresponding to transition n = 3 to n=1 is:

h = 6 \times 10^{- 34} J - s e c]

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