Important Questions For Practice
1. The work function of the following metal is given Na = 2.75 eV, K = 2.3 eV, Mo = 4.14 eV, Ni = 5.15 eV which of these metal will not give a photoelectric emission for radiation of wave length 3300 A0 from a laser
source placed at 1m away from the metal. What happens if the laser is brought nearer and placed 50 cm away.
2. Why photo-electrons ejected from a metal surface have different kinetic energies although the frequency of incident photons are same?
3. Define distance of the closest approach. An alpha-particle of kinetic energy тАШKтАЩ is bombarded on a thin gold foil. The distance of the closet approach is тАШrтАЩ. What will be the distance of closest approach for an alpha-particle of double the kinetic energy?
4. If the total number of neutrons and protons in a nuclear reaction is conserved how then is the energy absorbed or evolved in the reaction?
5. Particle of mass M at rest decays into two particles of masses m1 and m2 having velocities V1 and V2 respectively. Find the ratio of de-Broglie Wavelengths of the two particles.
Questions From Board Papers
1. What is the stopping potential applied to a photocell, in which electrons with a maximum kinetic energy of 5.6 eV are emitted.
2. If the amount of a radioactive substance is increased four times then how many times will the number of atoms disintegrating per unit time be increased?
3. Why does only a slow neutron (.03eV energy) cause the fission in the uranium nucleus and not the fast one?
4. In BohrтАЩs atomic model, the potential energy is negative and has a magnitude greater than the kinetic energy, what does this imply?
5. The half life of a radioactive element A is same as the mean life time of another radioactive element B. Initially, both have same number of atoms. B decay faster than A. Why?
Important Questions On 12th Physics Chapter 11
V is related to frequency by the equation
hv = eV
Where, e = Charge on an electron = 1.6 ├Ч 10^тИТ19 C and
h = PlanckтАЩs constant
Therefore,
h = e├ЧV/v
= 1.6├ЧуАЦ10уАЧ^(-19) ├Ч 4.12 ├ЧуАЦ10уАЧ^(-15)
= 6.592 ├ЧуАЦ10уАЧ^(-34) Js
Hence, the value of PlankтАЩs constant is 6.592├ЧуАЦ10уАЧ^(-34) Js.
тЗТ╬╗=h/p тАж (i)
Where, ╬╗ = Wavelength of the electromagnetic radiation
c = Speed of light
h = PlanckтАЩs constant
De Broglie wavelength of the photon is given as: ╬╗=h/mv
But p=mv, therefore
╬╗=h/p тАж (ii)
Where, m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.