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The de-Broglie wavelength of an electron moving with a velocity c/2 (c = velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is

1 : 4
1 : 2
1 : 1
2 : 1

182.
Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

$\frac{5}{27}$

$\frac{3}{23}$

$\frac{7}{29}$

$\frac{9}{31}$

$\frac{5}{27}$

$Wavelength for Lyman series λ_{L} = \frac{1}{R (1 - \frac{1}{4})} = \frac{4}{3 R} and wavelength for Balmer series λ_{B} = \frac{1}{R (\frac{1}{4} - \frac{1}{9})} = \frac{1}{R (\frac{5}{36})} = \frac{36}{5 R} ∴ \frac{λ_{L}}{λ_{B}} = \frac{4}{3 R} \times \frac{5 R}{36} = \frac{5}{27} \Rightarrow λ_{L} : λ_{B} = 5 : 27$

183.

The ionization energy of hydrogen is 13.6 eV. The energy of the photon released when an electron jumps from the first excited state (n = 2) to the ground state of a hydrogen atom is

3.4 eV
4.53 eV
10.2 eV
13.6 eV

184.

The de-Broglie wavelength of an electron is the same as that of a 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 0.5 MeV)

1 : 50
1 : 20
20 : 1
50 : 1

185.

Consider two particles of different masses. In which of the following situations the heavier of the two particles will have smaller de-Broglie wavelength ?