JEE Main Atom PYQ

51. (JEE Main 2013 (Offline))

In a hydrogen like atom electron make transition from an energy level with quantum number

n

to another with quantum number

(n - 1)

. If

n >> 1,

the frequency of radiation emitted is proportional to :

A. $\frac{1}{n}$

B. $\frac{1}{n^{2}}$

C. $\frac{1}{n^{\frac{3}{2}}}$

D. $\frac{1}{n^{3}}$

The correct answer is option (D)

$Δ$ E $= 13.6 Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

As $Δ$ E = h $υ$

$\Rightarrow$ $Δ$ E $\propto$ $υ$ $\propto$ $[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

Here n₁ = n - 1 and n₂ = n

$∴$ $υ$ $\propto$ $[\frac{1}{{(n - 1)}^{2}} - \frac{1}{n^{2}}]$

$\Rightarrow$ $υ$ $\propto$ $[\frac{2 n - 1}{n^{2} {(n - 1)}^{2}}]$

[ As $n >> 1,$ so we can assume n - 1 = n and 2n - 1 = 2n ]

$\Rightarrow$ $υ$ $\propto$ $[\frac{2 n}{n^{2} {(n)}^{2}}]$ $\propto$ $[\frac{2 n}{n^{4}}]$ $\propto$ $[\frac{2}{n^{3}}]$

52. (AIEEE 2012)

Hydrogen atom is excited from ground state to another state with principal quantum number equal to

4.

Then the number of spectral lines in the emission spectra will be :

A. 2

B. 3

C. 5

D. 6

The correct answer is option (D)

The possible number of the spectral lines is given

$= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

53. (AIEEE 2009)

The transition from the state

n = 4

to

n = 3

in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from :

A. $3 \to 2$

B. $4 \to 2$

C. $5 \to 4$

D. $2 \to 1$

The correct answer is option (C)

It is given that transition from the state $n = 4$ to $n = 3$ in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is $5$ $\to 4.$

54. (AIEEE 2007)

Which of the following transitions in hydrogen atoms emit photons of highest frequency ?

A. $n = 1$ to $n = 2$

B. $n = 2$ to $n = 6$

C. $n = 6$ to $n = 2$

D. $n = 2$ to $n = 1$

The correct answer is option (C)

We have no find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options $(c)$ and $(d)$ .

Frequency is given by

$h v = - 13.6 (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}})$

For transition from $n = 6$ to $n = 2,$

$v_{1} = \frac{- 13.6}{h} (\frac{1}{6^{2}} - \frac{1}{2^{2}})$

$= \frac{2}{9} \times (\frac{13.6}{h})$

For transition from $n = 2$ to $n = 1,$

$v_{2} = \frac{- 13.6}{h} (\frac{1}{2^{2}} - \frac{1}{1^{2}})$

$= \frac{3}{4} \times (\frac{13.6}{h}) .$

$∴$ $v_{1} > v_{2}$

55. (AIEEE 2005)

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

AIEEE 2005 Physics - Atoms and Nuclei Question 179 English

AIEEE 2005 Physics - Atoms and Nuclei Question 179 English

A. $i v$

B. $i i i$

C. $i i$

D. $i$

The correct answer is option (B)

KEY CONCEPT : $E = R h c [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}]$

$E$ will be maximum for the transition for which

$[\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}]$ is maximum. Here $n_{2}$ is the higher energy level

Clearly, $[\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}]$ is maximum for the third

transition, i.e. $2 \to 1.$ $I$ transition represents the absorption of energy.