A source S₁ is producing, 10¹⁵ photons/s of wavelength 5000 $\stackrel{\circ }{A}$. Another source S₂ is producing 1.02$\times$ 10¹⁵ photons per second of wavelength 5100 $\stackrel{\circ }{A}$. Then, (power of S₂)/(power of S₁) is equal to

1. 1.00                 2. 1.02

3. 1.04                 4. 0.98

The electron in the hydrogen atom jumps from excited state $\left(n=3\right)$ to its ground state $\left(n=1\right)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \mathrm{eV},$ the stopping potential is estimated to be (the energy of the electron in the nth state ${\mathrm{E}}_{\mathrm{n}}=-\frac{13.6}{{\mathrm{n}}^2}\mathrm{eV}$)

1. $5.1 \mathrm{V}$                                               

2. $12.1 \mathrm{V}$

3. $17.2 \mathrm{V}$                                             

4. $7 \mathrm{V}$

The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three difference radiations. Which one of the following is a correct statement?

(1) Curves a and b represent incident radiations of different frequencies and different intensities

(2) Curves a and b represent incident radiations of same frequency but of different intensities 

(3) Curves b and c represent incident radiations of different frequencies and different intensities

(4) Curves b and c represent incident radiations of same frequency having same intensity 

The number of photoelectrons emitted for light of a frequency v (higher than the threshold frequency $v_0$) is proportional to

1. $v-v_0$

2. threshold frequency $\left(v_0\right)$

3. intensity of light

4. frequency of light (v)

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5V lies in the 

(1) ultraviolet region                               

(2) visible region

(3) infrared region                                   

(4) X-ray region 

An electron $\left(mass m\right)$ with an initial velocity v=${\mathrm{v}}_0\hat{\mathrm{i}} \left({\mathrm{v}}_0>0\right)$ is in an electric field E$=-E_0 \hat{i}\left(E_0=cons\tan t > 0\right).$ It's de Broglie wavelength at the time $t$ is given by:

(1) $\frac{{\lambda }_0}{\left(1+{\displaystyle \frac{e{E}_0}{m}}{\displaystyle \frac{t}{v_0}}\right)}$

(2) ${\lambda }_0\left(1+\frac{e{E}_{0}t}{m{v}_0}\right)$

(3) ${\lambda }_0$

(4)${\lambda }_{0}t.$