Which of the following transitions in hydrogen atoms emit photons of highest frequency?

• n = 2 to n = 6

• n = 6 to n = 2

• n = 2 to n =1

• n = 2 to n =1

The energy spectrum of β-particles [number N(E) as a function of β-energy E] emitted from a radioactive source is

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

• lll

• IV

• I

• I

An electron from various excited states of hydrogen atom emits radiation to come to the ground state. Let λ_n, λ_g, be the de Broglie wavelength of the electron in the n^th state and the ground state respectively. Let Λ_n be the wavelength of the emitted photon in the transition from the n^th state to the ground state. For large n, (A, B are constants)

• ${\Lambda }_n^2 \approx \lambda$

• ${\Lambda }_n \approx A + \frac{B}{{\lambda }_n^2}$

• ${\Lambda }_n \approx A + B{\lambda }_n$

• ${\Lambda }_n^2 \approx A + B{\lambda }_n^2$

If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is:

• v_L/25

• 25v_L

• 16 v_L

• v_L/16

16.

What wavelength must electromagnetic radiation have if a photon in the beam has the same momentum as an electron moving with a speed 1.1 x 10⁵ m/s (Planck's constant = 6.6 x 10^-34 J-s, rest mass of electron = 9 x 10^-31 kg ?

• 2/3 nm

• 20/3 nm

• 4/3 nm

• 40/3 nm

An electron, a neutron and an alpha particle have same kinetic energy and their de-Broglie wavelength are λ_e, λ_n, λ_α and respectively. Which statement is correct about their de-Broglie wavelengths ?

• λ_e > λ_n > λ_α

• λ_e < λ_n > λ_α

• λ_e < λ_n < λ_α

• λ_e > λ_n < λ_α

An electron of mass m_e and a proton of mass m_p are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is

• 1

• $\sqrt{\frac{{\mathrm{m}}_{\mathrm{e}}}{{\mathrm{m}}_{\mathrm{p}}}}$

• $\sqrt{\frac{{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{e}}}}$

• $\frac{{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{e}}}$

The significant result deduced from the Rutherford's scattering experiment is that

• whole of the positive charge is concentrated at the centre of atom

• there are neutrons inside the nucleus

• α-particles are helium nuclei

• electrons are embedded in the atom

The de-Broglie wavelength and kinetic energy of a particle is 2000 $\stackrel{\circ }{\mathrm{A}}$ and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de-Broglie wavelength is

• $2 \stackrel{\circ }{\mathrm{A}}$

• $1 \stackrel{\circ }{\mathrm{A}}$

• $4 \stackrel{\circ }{\mathrm{A}}$

• $10 \stackrel{\circ }{\mathrm{A}}$