If the kinetic energy of the particle is increased to \(16\) times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
1. \(25\)
2. \(75\)
3. \(60\)
4. \(50\)

For photoelectric emission from certain metals, the cutoff frequency is \(\nu.\) If radiation of frequency \(2\nu\)$\mathrm{}$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
(\(m\) is the electron mass)

A \(200~\text{W}\) sodium street lamp emits yellow light of wavelength \(0.6~\mu\text{m}.\)$$ Assuming it to be \(25\%\) efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:
1. \(1.5\times 10^{20}\)
2. \(6\times 10^{18}\)
3. \(62\times 10^{20}\)
4. \(3\times 10^{19}\)

An \(\alpha\text-\)particle moves in a circular path of radius \(0.83~\text{cm}\) in the presence of a magnetic field of \(0.25~\text{Wb/m}^2.\) The de-Broglie wavelength associated with the particle will be:
1. \(1~\mathring{A}\)
2. \(0.1~\mathring{A}\)
3. \(10~\mathring{A}\)
4. \(0.01~\mathring{A}\)

A radioactive nucleus of mass M emits a photon of frequency $\nu$ and the nucleus will recoil. The recoil energy will be:

1.  $\frac{{\mathrm{h}}^2{\mathrm{\nu }}^2}{2{\mathrm{Mc}}^2}$

2.  zero

3.  $\frac{\mathrm{h\nu }}{\mathrm{c}\sqrt{2\mathrm{M}}}$

4.  $\frac{\mathrm{c}\sqrt{2\mathrm{M}}}{\mathrm{h\nu }}$

A source S₁ is producing 10¹⁵ photons per sec of wavelength 5000 Å. Another source S₂ is producing 1.02×10¹⁵ photons per second of wavelength 5100 Å. Then, (power of S₂)/(power of S₁) is equal to:

1. 1.00

2. 1.02

3. 1.04

4. 0.98