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Standard XII

Chemistry

Emission and Absorption Spectra

For Li2+, w...

Question

For ${L i}^{2 +}$ , when an electron falls from a higher orbit to $n$ th orbit, all the three types of lines, i.e., Lyman, Balmer and Paschen was found in the spectrum. Here, the value of ' $n$ ' will be ___.

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Solution

The correct option is A $1$
$\frac{1}{λ} = {R Z}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
given, $n_{1} = n$
lyman series is obsereved only if $n_{1} = n = 1$

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Similar questions

Q. Total number of lines in Lyman series of H spectrum will be (where n = no. of orbits)

Q. Number of visible lines when an electron returns from

5^{t h}

orbit to

2^{n d}

orbit in H-spectrum:

Q. Assertion :In

L i^{2 +}

sample, if an electron makes transition from higher state to n=2, then the photon observed will fall in the visible range. Reason: Line falling in n=2 is Balmer series line may belong to visible range H-atom.

In Balmer series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of th e following inner orbit jumps of the electron for bohr orbits in an atom of hydrogen?

A.3->2

B. 5->2

C. 4->1

D. 2->5

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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For Li2+, when an electron falls from a higher orbit to nth orbit, all the three types of lines, i.e., Lyman, Balmer and Paschen was found in the spectrum. Here, the value of 'n' will be ___.

The correct option is A 11λ=RZ2(1n21−1n22)given,n1=nlyman series is obsereved only if n1=n=1

For ${L i}^{2 +}$ , when an electron falls from a higher orbit to $n$ th orbit, all the three types of lines, i.e., Lyman, Balmer and Paschen was found in the spectrum. Here, the value of ' $n$ ' will be ___.

The correct option is A $1$
$\frac{1}{λ} = {R Z}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
given, $n_{1} = n$
lyman series is obsereved only if $n_{1} = n = 1$