Atoms

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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

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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.

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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.

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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.

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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= $\frac{p{R}_0^{\delta }}{m{r}^{1+\delta }}$

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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 $*2=27.2eV$

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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