Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- The transition of an electron from a higher energy to a lower energy level can appears in the form of electromagnetic radiation because electrons interact only electromagnetically.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- On removing one electron from He4 and He3, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from He4 and He3 atoms contain one electron and are hydrogen like atoms.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- the reason behind this is the binding energy or we can say the mass defect.

BE= mass defect (931MeV ). So whatever energy is liberated in fission and fusion changes the mass slighthly. That why there is a defect in mass

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- when r0=1A^o

Let $\epsilon =2+\delta$

F= $\frac{{kq}_1{q}_2}{r^{2+\delta }}{R}_0^{\delta }$

Where $\frac{{kq}_1{q}_2}{r^2}$ = ( ${1.6*{10}^{-19})}^2*9*{10}^9=23.04*{10}^{-29}N/m2$

Let p= $23.04*{10}^{-29}N/m2$

Electrostatic force is balanced by centripetal force

Mv²/r or v²= $\frac{p{R}_0^{\delta }}{m{r}^{1+\delta }}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- m_p= 10^-6

M_pc²=10^{-6 $*electronmass*c^2$}

 = 0.8 $*{10}^{-19}J$

Wavelength associated with both of them is same

$\frac{h}{m_{p}c}=\frac{hc}{m_p{c}^2}=\frac{{10}^{-34}*3*{10}^8}{0.8*{10}^{-19}}=4*{10}^{-7}m$

U(r)=- $\frac{e^2}{4\pi {\epsilon }_o}\frac{\mathrm{e}\mathrm{x}\mathrm{p}?(-\lambda t)}{r}$

Mvr=h

V=h/mr

Mv²/r=( $\frac{e^2}{4\pi {\epsilon }_o}$ ) ( $\frac{1}{r^2}+\frac{\lambda }{r}$ )

$\frac{h^2}{m{r}^3}=\left(\frac{e^2}{4\pi {\epsilon }_o}\right)\left(\frac{1}{r^2}-\frac{\lambda }{r}\right)$

$\frac{h^2}{m}=\frac{e^2}{4\pi {\epsilon }_o}{r}_B$

$kinecticwillbedoubleofgroundstateso$

13.6 $*2=27.2eV$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- the energy of nth state E_n=-Z²R $\frac{1}{n^2}$  where R is constant and Z=24

The energy release ina transition from 2 to 1 $?E=Z^{2}R\left(1-\frac{1}{4}\right)=\frac{3}{4}{Z}^{2}R$

The energy required to eject a n=4 electron is E₄= Z²R1/16

So kinetic energy of auger electron is, KE= Z²R (3/4-1/16)=1/16Z²R

= $\frac{11}{16}*24*24*13.6eV=5385.6eV$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- in this type of situation centripetal force is balanced by electrostatic force

So mv²/r= -ke²/r²

According to bohr postuates

Mvr = h when n=1

On solving n  $\frac{h^2}{m^2{r}^2}\frac{1}{r}={ke}^2/r^2$

So after solving r= 0.51A⁰

So potential energy  $\frac{{ke}^2}{r}$ = -27.2eV , KE= mv²/2

= $\frac{1}{2}m\frac{h^2}{r^2}=\frac{h}{2m{r}^2}=+13.6eV$

But if R

If R>>r the electron moves inside the sphere with radius r'

Charge inside r'⁴=e $\frac{{r\text{'}}^3}{R^3}$

So r' = $\frac{h^2}{m}\frac{k}{e^2}\frac{R^3}{{r\text{'}}^3}$

So r'⁴= 0.51A⁰R³

= 510(A⁰)⁴

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- total energy of electron

E= $\frac{\mu Z^2{e}^4}{8{\epsilon }_o^2{h}^2}\frac{1}{n^2}$

The frequency hv= $\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\left(1-\frac{1}{4}\right)=\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\frac{3}{4}$

$?\lambda$ = ${\lambda }_D-{\lambda }_H$

$100*\frac{?\lambda }{{\lambda }_H}$ = $\frac{{\lambda }_D-{\lambda }_H}{{\lambda }_H}*100$ = $\frac{{\mu }_D-{\mu }_H}{{\mu }_H}*100$

= $\frac{\frac{m_{e{M}_D}}{m_e+M_H}-\frac{m_{e{M}_D}}{m_e-M_H}}{\frac{m_{e{M}_D}}{m_e+M_H}}*100$

When m_e<H<

after solving we get 2.714 $*{10}^{-2}\%$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- as we know total energy in stationary orbit is

E_n=- $\frac{{-me}^4}{8n^2{{\epsilon }_0}^2{h}^2}$  where sign have usual meaning.

According to bohr third postulate h $\nu =E_{f-}{E}_i$

$\nu =$ - $\frac{{me}^4}{8{{\epsilon }_0}^2{h}^3}(\frac{1}{n_f^1}-\frac{1}{n_i^2})$

$\lambda \propto \frac{1}{\propto }$

Where $\propto$  is the reduced mass

Reduced mass for H=_H=;m_e(1-m_e/M)

D= _D; m_e(1-m_e/2M)

 =m_e(1-m_e/2M)(1+m_e/2M)

If for hydrogen deuterium, the wavelength

$\frac{{\lambda }_D}{{\lambda }_H}=\frac{{\propto }_H}{{\propto }_D}$ = (1+ $\frac{m_e}{2M}$ )^-1= (1- $\frac{1}{2*1840}$ )

${\lambda }_D={\lambda }_H*0.99973$  so lines emitted are 1217.7A⁰,1027.7A⁰,974.04A⁰