Question

The threshold frequency for a metallic surface corresponds to an energy of $$6.2\,eV$$  and the stopping potential for a radiation incident on this surface is $$5\,V.$$  The incident radiation lies in

A. ultra-violet region  
B. infra-red region
C. visible region
D. X-ray region
Answer :   ultra-violet region
Solution :
$$\eqalign{ & \phi = 6.2eV = 6.2 \times 1.6 \times {10^{ - 19}}J \cr & V = 5volt,\frac{{hc}}{\lambda } - \phi = e{V_0} \cr & \Rightarrow \lambda = \frac{{hc}}{{\phi + e{V_0}}} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}}\left( {6.2 + 5} \right)}} \approx {10^{ - 7}}m \cr} $$

Releted MCQ Question on
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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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