Physics

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

Explanation- The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius (a_n) . Let the velocity of electron is v

∴ Number of revolutions per unit time= $\frac{2\pi a_0}{V}$

The electric current is given by i=q/t, if q charge flows in time t. Here,   q=e

The electric current is given by i= $\frac{2\pi a_0}{V}e$

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

Explanation- For a He -nucleus with charge 2e and electrons of charge− e, the energy level in ground state is

E=-13.6Z²/n²= -54.4eV

Thus, the ground state will have two electrons each of energy E and the total ground state

energy would be   - (4*13.6)eV = -54.4eV.

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

Explanation- The total energy of the electron in the stationary states of the hydrogen atom is given by E_n = - $\frac{m{e}^4}{8n^2{\epsilon }_0^2{h}^2}$

where signs are as usual and the m that occurs in the Bohr formula is the reduced mass of

electron and proton. Also, the total energy of the electron in the ground state of the

hydrogen atom is−13.6 eV. For H-atom reduced mass m_e . Whereas for positronium, the

reduced mass is m=m_e/2

Hence, the total energy of the electron in the ground state of the positronium atom is

 -13.6/2=-6.8eV

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

Explanation- According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as

L=nh/2 $\pi L\propto$ n

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

Explanation- If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

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 $*2=27.2eV$