Question

Two identical photocathodes receive light of frequencies $${f_1}$$ and $${f_2}.$$ If the velocites of the photo electrons (of mass $$m$$) coming out are respectively $${v_1}$$ and $${v_2},$$ then

A. $$v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)$$  
B. $${v_1} + {v_2} = {\left[ {\frac{{2h}}{m}\left( {{f_1} + {f_2}} \right)} \right]^{\frac{1}{2}}}$$
C. $$v_1^2 + v_2^2 = \frac{{2h}}{m}\left( {{f_1} + {f_2}} \right)$$
D. $${v_1} - {v_2} = {\left[ {\frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)} \right]^{\frac{1}{2}}}$$
Answer :   $$v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)$$
Solution :
For one photocathode
$$h{f_1} - W = \frac{1}{2}mv_1^2\,......\left( {\text{i}} \right)$$
For another photo cathode
$$h{f_2} - W = \frac{1}{2}mv_2^2\,......\left( {{\text{ii}}} \right)$$
Subtracting (ii) from (i) we get
$$\eqalign{ & \left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \cr & \therefore h\left( {{f_1} - {f_2}} \right) = \frac{m}{2}\left( {v_1^2 - v_2^2} \right) \cr & \therefore v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right) \cr} $$

Releted MCQ Question on
Modern Physics >> Modern Physics Miscellaneous

Releted Question 1

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $$6\,eV$$  fall on it is $$4\,eV.$$  The stopping potential, in volt, is

A. 2
B. 4
C. 6
D. 10
Releted Question 2

Electrons with energy $$80\,keV$$  are incident on the tungsten target of an X-ray tube. $$K$$-shell electrons of tungsten have $$72.5\,keV$$  energy. X-rays emitted by the tube contain only

A. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$
B. a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths
C. the characteristic X-ray spectrum of tungsten
D. a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of $$0.155\mathop {\text{A}}\limits^ \circ $$  and the characteristic X-ray spectrum of tungsten.
Releted Question 3

The intensity of X-rays from a Coolidge tube is plotted against wavelength $$\lambda $$ as shown in the figure. The minimum wavelength found is $${\lambda _C}$$ and the wavelength of the $${K_\alpha }$$ line is $${\lambda _K}.$$  As the accelerating voltage is increased
Modern Physics Miscellaneous mcq question image

A. $${\lambda _K} - {\lambda _C}$$   increases
B. $${\lambda _K} - {\lambda _C}$$   decreases
C. $${\lambda _K}$$ increases
D. $${\lambda _K}$$ decreases
Releted Question 4

The potential difference applied to an X-ray tube is $$5k\,V$$  and the current through it is 3.2$$mA.$$  Then the number of electrons striking the target per second is

A. $$2 \times {10^{16}}$$
B. $$5 \times {10^{6}}$$
C. $$1 \times {10^{17}}$$
D. $$4 \times {10^{15}}$$

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