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 the lowest orbit while Balmer series is formed when the electron returns to the second orbit- Similarly- Paschen- Brackett- and Pfund series are formed when an electron returns to the third- fourth- and fifth from higher orbits- respectively-A maximum number of lines produced when an electron jumps from nth level to ground level is equal to nn - 1-2- If the electron comes back from the energy level having energy E2 to the energy level having energy E1- then difference may be expressed in terms of energy of photon as - E2 - E1 - - E- - - hc- E-Since h and c are constant- -E corresponds to definite energy thus- each transition from one energy level to another will produce a light of definite wavelength- This is actually observed as a line in the spectrum of a hydrogen atom-Wave number of line is given by the formula - v - RZ2 1-n12 - 1-n22Where R is a Rydberg constant-In a single isolated atom- an electron makes transition from a fifth excited state to second state then maximum number of different types of photons observed is -

The correct option is B 6As we know,No of spectral lines = (n_2 n_1)(n_2 n_1 +1)2 n_2 = 5^th exited state = 6 n_1 = 3 2nd excited stateNo of ...

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Byju's Answer

Standard XII

Chemistry

Emission and Absorption Spectra

The only elec...

Question

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 the lowest orbit while Balmer series is formed when the electron returns to the second orbit. Similarly, Paschen, Brackett, and Pfund series are formed when an electron returns to the third, fourth, and fifth from higher orbits, respectively.
A maximum number of lines produced when an electron jumps from nth level to ground level is equal to $\frac{n (n - 1)}{2}$ .
If the electron comes back from the energy level having energy $E_{2}$ to the energy level having energy $E_{1}$ , then difference may be expressed in terms of energy of photon as :
$E_{2} - E_{1}$ = $Δ E$ , $λ = \frac{h c}{Δ E}$ .
Since $h$ and $c$ are constant, $Δ$ E corresponds to definite energy thus, each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of a hydrogen atom.
Wave number of line is given by the formula :
$v$ = $R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where $R$ is a Rydberg constant.

In a single isolated atom, an electron makes transition from a fifth excited state to second state then maximum number of different types of photons observed is :

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Solution

The correct option is B 6
As we know,
No of spectral lines $= \frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2}$
$n_{2} = 5^{t h}$ exited state = 6
$n_{1} = 3$ 2nd excited state

No of lines $= \frac{(6 - 3) (6 - 3 + 1)}{2} = 6$
Option C is correct.

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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]

Q. When energy is supplied to an atom, the electrons of the atom move to higher energy orbits depending on the amount of energy absorbed. When these electrons return to any of the lower energy orbits, they emit energy. Depending on whether the electrons return to the 1st, 2nd, 3rd, 4th or 5th orbits, the series formed are called Lyman, Balmer, Paschen, Brackett and Pfund series respectively.
If the electron comes back from the energy level

E_{2}

to the energy level

E_{1}

, then the difference may be expressed in terms of the energy of photon as

E_{2} - E_{1} = Δ E

λ = \frac{h c}{Δ E}

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.
Wave number of the line is given by the formula

- ν = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

Where

R = 1.097 \times 10^{5} c m^{- 1}

is Rydberg constant.

\begin{matrix} List - I & List - II (I) If the ionisation potential for hydrogen-like atom & (P) 5.48 \times 10^{- 32} c R J in a sample is 122.0 e V, then the energy of the last line of Paschen series for this atom is: (II) In a single isolated atom, an electron makes & (Q) 6.6 \times 10^{- 32} c R J transition from fifth excited state such that no spectral lines are observed in the visible range then maximum number of different types of photons observed is: (III) The difference in energy of the second line & (R) 6 of Lyman series and last line of Brackett series in a hydrogen sample is: (IV) The energy of the electromagnetic radiation & (S) 4 emitted during the transition of electron in between the two levels of L i^{2 +} ion whose principal quantum number sum is 4 and difference is 2 is: (T) 6.6 \times 10^{- 34} c R J (U) 0.53 \times 10^{- 30} c R J \end{matrix}

Which of the following options has the correct combination considering List-I and List-II?

Q. Find the number of spectral line in Paschen series emitted by atomic

H

, when electron excited from ground state to

70^{t h}

energy level return back.

Q. (a) As long as an electron revolves in a particular orbit, the electron does not lose its energy. Therefore, these orbits are called stationary orbits and the electrons are said to be in stationary energy states.
(b) Each orbit or shell is associated with a definite amount of energy. Hence these are also called energy levels.
(c) An electron jumps from a lower energy level to a higher energy level, by absorbing energy. It jumps from a higher energy level to a lower energy level, by emitting energy in the form of electromagnetic radiation.
(d) Electrons move around the nucleus in specified circular paths called orbits or shells or energy levels and are designated as K, L, M, N shells respectively.

Arrange the above postulates of Bohr's theory in a correct sequence:

Q. When electronic transition occurs from higher energy state to a lower energy state with energy difference equal to

Δ E

expressed in electron volts, the wavelength of line emitted is approximately equal to:

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The correct option is B 6As we know,No of spectral lines =(n2−n1)(n2−n1+1)2n2=5th exited state = 6n1=3 2nd excited stateNo of lines =(6−3)(6−3+1)2=6Option C is correct.

The correct option is B 6
As we know,
No of spectral lines $= \frac{(n_{2} - n_{1}) (n_{2} - n_{1} + 1)}{2}$
$n_{2} = 5^{t h}$ exited state = 6
$n_{1} = 3$ 2nd excited state

No of lines $= \frac{(6 - 3) (6 - 3 + 1)}{2} = 6$
Option C is correct.