Physics

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

Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.

Given, E_max= hv- ${?}_0$

Now, maximum energy for emitted electrons is E_max= h2v- ${?}_0=2\mathrm{h}\mathrm{v}-{?}_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible

This is a short answer type question as classified in NCERT Exemplar

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

By lens makers formula 1/f= $(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

So f_br so focal length of blue is less than red light 

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

Explanation- $\lambda =h/\sqrt[]{2mqv}$

$\lambda \propto \frac{1}{\sqrt[]{mq}}$

$\frac{{\lambda }_p}{{\lambda }_a}=\frac{\sqrt[]{m_a{q}_a}}{\sqrt[]{m_p{q}_p}}=\frac{\sqrt[]{4m_p*2e}}{\sqrt[]{m_p*e}}$ = $\sqrt[]{8}$

So wavelength of proton is $\sqrt[]{8}$ times of alpha particle

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

Explanation- from conservation of momentum

P_c=P_A+P_B

= $\frac{h}{{\lambda }_c}=\frac{h}{{\lambda }_A}+\frac{h}{{\lambda }_B}$

$\frac{h}{{\lambda }_c}=\frac{h{\lambda }_A+h{\lambda }_B}{{\lambda }_A{\lambda }_B}$

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 1 when both momentum are positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 2 when both momentum are negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 3 when 1 is negative and 2 is positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_B-{\lambda }_A}$

Case4 when 1 is positive and 2 is negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A-{\lambda }_B}$

This is a long answer type question as classified in NCERT Exemplar

Time from S to P₁ is

t₁ = $\frac{S{P}_1}{c}=\frac{\sqrt[]{u^2+b^2}}{c}$

$\frac{u}{c}(1+\frac{1}{2}\frac{b^2}{u^2})$

Time from P₁ to O is

t₂ = $\frac{P_{1}O}{c}=\frac{\sqrt[]{v^2+b^2}}{c}$ ; $\frac{v}{c}\left(1+\frac{1}{2}\frac{b^2}{v^2}\right)$

the time required travel through lens is t₁ = $\frac{(n-1)wb}{c}$

so total time is t=1/c(u+v+b²/2D+(n-1)(wo+b²/ $\propto$ ))

after solving we get $\propto =2\left(n-1\right)D$

differentiating with respect to time

t=1/c(u+v+b²/2D+(n-1)K1In(K₂b))

dt/db=0=b/D-(n-1)K₁/b

b²= (n-1)K1D

b= $\sqrt[]{(\mathrm{n}-1)K_1\mathrm{D}}$

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

Explanation- Suppose nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.

N_A=2n_B

energy of falling photon A=hv_A

B=hv_B

as we know intensities are same

n_Ahv_A=n_Bhv_B

v_a/v_b=n_B/n_A=1/2

v_B=2v_A

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

Explanation-

$?p=\frac{h}{?x}=\frac{h}{2\pi ?x}$

= $\frac{6.64*{10}^{-34}}{2*22/7*{10}^{-9}}=1.05*{10}^{-25}$ kgm/s

E= p²/2m= $\frac{{(1.05*{10}^{-25})}^2}{2*9.1*{10}^{-31}}$ J= 3.8 $*{10}^{-2}$ eV

This is a long answer type question as classified in NCERT Exemplar

Let us consider a portion of a ray between x and x+dx inside the liquid. Let the angle of incidence at x be θ and let it enter the thin column at height y. Because of the bending it shall emerge at x+dx  with an angle θ+dθ and at a height y+dy . From Snell's law,

$\mu ysin\theta =\mu (y+dy)sin(\theta +d\theta )$

$\mu ysin\theta \left(\mu y+\frac{d\mu }{dy}dy\right)sin\theta cosd\theta +cos\theta si$ nd $\theta$

$\mu ycos\theta d\theta$ $\frac{d\mu }{dy}dysin\theta$

d $\theta -\frac{d\mu }{dy}dytan\theta$

tan $\theta =\frac{dx}{dy}$

$\theta =-\frac{1d\mu d}{\mu dy}$