AtomsNEET Physics · Class 12 · NCERT Chapter 12

Solution

The energy of the photon is given by $E=\frac{hc}{\lambda }$. Substituting $\lambda =975\times 10^{-10}\text{ m}$, $E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{975\times 10^{-10}}\approx 20.4\text{ eV}$. This energy corresponds to the transition from the ground state (n=1) to the n=4 level. The number of spectral lines emitted when an electron de-excites from n=4 to lower levels is given by $\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$. Therefore, option (c) is correct.

Solution

1 λ1 =Re(1 12− 1 22) 1 λ2 =Re(1 22− 1 32) λ1 λ2 = 5 27

Solution

For last Balmer series 22 11 1 2b R ⎡⎤=- ⎢⎥λ ∞⎣⎦ 4 b Rλ= For last Lyman series 22 11 1 1l R ⎡⎤=- ⎢⎥λ ∞⎣⎦ www.vedantu.com 4 1 l Rλ= 4 1 b l R R λ =λ 4b l λ =λ

Solution

In the Bohr model, the total energy $E$ of an electron in a hydrogen atom is given by $E=-\frac{13.6}{n^2}\text{eV}$, and the kinetic energy $K$ is $K=\frac{13.6}{n^2}\text{eV}$. The ratio of kinetic energy to total energy is $\frac{K}{E}=\frac{\frac{13.6}{n^2}}{-\frac{13.6}{n^2}}=-1$, so the ratio is $1:-2$. Option (d) is correct.

Solution

Work done = ∫(20+10𝑦)𝑑𝑦 =25 𝐽

Solution

K = − E = + 3.4 eV U = 2E = − 6.8 eV

Solution

0 2=n EE n , For first excited state ⇒ n = 2 For second excited state ⇒ n = 3 ⇒ 0 1 02 94 4 9 = = E T ET SECTION-B

Solution

The shortest wavelength in the Balmer series corresponds to the transition from $n=\infty$ to $n=2$. For the Brackett series, the shortest wavelength corresponds to the transition from $n=\infty$ to $n=4$. Using the Rydberg formula, $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, the ratio of the wavelengths is $\frac{{\lambda }_{\text{Brackett}}}{{\lambda }_{\text{Balmer}}}=\frac{2^2}{4^2}=\frac{1}{4}\Rightarrow {\lambda }_{\text{Brackett}}=4{\lambda }_{\text{Balmer}}$. Therefore, option (c) is correct.

Solution

Energy difference ΔE = hc λ ∴ 1 Eλ∝ Δ 6 2 5 2 4 2 3 2( ) ( ) ( ) ( )E E E E− − − −Δ > Δ > Δ > Δ 6 2 5 2 4 2 3 2− − − −λ < λ < λ < λ A-III, B-IV, C-II, D-I

Solution

2 nH 2 ZE R J n ⎛⎞=− ⎜⎟⎜⎟⎝⎠ For He+ (n = 1), 2 n H H 2 2E x R 4R 1 ⎛⎞= − = − = − ⎜⎟⎜⎟⎝⎠ ∴ H xR 4= For Be3+ (n = 2), 2 nH 2 ZE R J n ⎛⎞=− ⎜⎟⎜⎟⎝⎠ x 4 4 xJ4 2 2 ×⎛⎞= − × = −⎜⎟ ×⎝⎠

Solution

Sol.
Given, force is constant

$F=\frac{m{v}^2}{r}$

$\Rightarrow \frac{v^2}{r}=\text{constant}$

$\Rightarrow r\propto v^2...(1)$

$\&L=mvr=\frac{nh}{2\pi }...(2)$

$\Rightarrow \text{on solving equation (1) and equation (2)}$

$v\propto n^{1/3}\text{ and }r\propto n^{2/3}$

Solution

Sol. $r=0.052n^2$
For $n=2$
$r=0.052\times 4$
$=0.208\text{nm}$
$Mvr=\frac{nh}{2\pi }$
$\lambda =\frac{h}{Mv}=\pi r$
$=3.14\times 0.208\text{nm}$
$=0.65317\text{nm}$
$\approx 0.67\text{nm}$

Solution

$E_n=\frac{-2.18\times 10^{-18}\cdot z^2}{n^2}\text{J};r_n=\frac{52.9\cdot n^2}{z}\text{pm}$

For He⁺ (Z = 2, n = 1):

$E_{{\text{He}}^{+}}=-2.18\times 10^{-18}\times 4=-8.72\times 10^{-18}\text{J}$

$r_{{\text{He}}^{+}}=\frac{52.9\times 1}{2}=26.45\text{pm}$

For Li²⁺ (Z = 3, n = 1):

$E_{{\text{Li}}^{2+}}=-2.18\times 10^{-18}\times 9=-19.62\times 10^{-18}\text{J}$

$r_{{\text{Li}}^{2+}}=\frac{52.9\times 1}{3}=17.63\text{pm}$