What is the de-Broglie wavelength associated with an electron accelerated through a voltage of \(900\) V?
1. \(0.31~\mathring{A}\)
2. \(0.41~\mathring{A}\)
3. \(0.5~\mathring{A}\)
4. \(0.16~\mathring{A}\)

The maximum kinetic energy of photoelectron emitted from the surface of work function f due to incidence of light of frequency n is E. If the frequency of incident light is doubled, then maximum kinetic of emitted photon will be

1.  2E

2.  2E - f

3.  2E + f

4.  2E + 2f

If V be the accelerating voltage, then the maximum frequency of continuous x-rays is given by

1.  $\frac{eh}{V}$

2.  $\frac{hV}{e}$

3.  $\frac{eV}{h}$

4. $\frac{h}{eV}$

When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2$\lambda$, the stopping potential is $\frac{V}{4}$ .The threshold wavelength for metallic surface is:

(a) 5$\lambda$                (b)$\frac{5}{2}$$\lambda$

(c) 3$\lambda$                (d) 4$\lambda$

An electron of mass m and a photon have the same energy E. Find the ratio of de-Broglie wavelength associated with the electron to that associated with the photon. (c is the velocity of light)

^{$1.$ ${\left(\frac{E}{2m}\right)}^{1/2}$
$$
$2.$ $c{\left(2mE\right)}^{1/2}$
$$
$3.$ $\frac{1}{c}{\left(\frac{2m}{E}\right)}^{1/2}$
$$
$4.$ $\frac{1}{c}{\left(\frac{E}{2m}\right)}^{1/2}$}
 

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (c=velocity of light)

1. E/c

2. 2E/c

3. 2E/c²

4. E/c²