An electron, a doubly ionized helium ion (He⁺⁺) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $$\lambda $$_e, $$\lambda $$_He⁺⁺ and $$\lambda $$_p is :

In a photoelectric effect experiment, the graph of stopping potential V versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased :

Particle A of mass m_A = $${m \over 2}$$ moving along the x-axis with velocity v₀ collides elastically with another particle B at rest having mass m_B = $${m \over 3}$$. If both particles move along the x-axis after the collision, the change $$\Delta $$$$\lambda $$ in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength ($$\lambda $$₀) before collision is :

Given figure shows few data points in a phot electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is:
(Plancks constant h = 6.62 × 10^–34 J.s)