An electron and an alpha particle are accelerated by the same potential difference. Let $$\lambda_e$$ and $$\lambda_\alpha$$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

If $$\phi$$ is the work function of photosensitive material in $$\mathrm{eV}$$ and light of wavelength of numerical value $$\lambda=\frac{h c}{e}$$ metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take $$h$$-Plank's constant, $$c$$-velocity of light in free space) is (in SI units):

The graph which shows the variation of $$\left(\frac{1}{\lambda^2}\right)$$ and its kinetic energy, $$E$$ is (where $$\lambda$$ is de Broglie wavelength of a free particle):

If $$c$$ is the velocity of light in free space, the correct statements about photon among the following are:

A. The energy of a photon is $$E=h v$$.

B. The velocity of a photon is $$c$$.

C. The momentum of a photon, $$p=\frac{h v}{c}$$.

D. In a photon-electron collision, both total energy and total momentum are conserved.

E. Photon possesses positive charge.

Choose the correct answer from the options given below: