When electron jumps from n = 4 to n = 2 orbit, we get
Second line of Lyman series
Second line of Balmer series
Second line of Paschen series
An absorption line of Balmer series
An α-particle of energy 5 MeV is scattered through 180o by a fixed Uranium nucleus. The distance of the closest approach is of the order of
1 Å
10-10 cm
10-12 cm
10-15 cm
If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is
102 eV
Zero
3.4 eV
6.8 eV
The ionization energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is
3.40 eV
1.51 eV
0.85 eV
0.66 eV
In terms of Bohr radius a0, the radius of the second Bohr orbit of hydrogen atom is given by
4 a0
8 a0
√2 a0
2 a0
The ionization energy of hydrogen atoms is 13.6 eV, the ionization energy of helium atom would be
13.6 eV
27. 2 eV
54. 4 eV
In the Bohr model of the hydrogen atom, let R, V and E represent the radius of the orbit, the speed of electron and the total energy of the electron respectively.Which of the following quantity is proportional to the quantum number n?
E/V
R/E
VR
If the electron in a hydrogen atom jumps from an orbit with level n2 = 3 to an orbit with level n1 = 2, the emitted radiation has a wavelength given by
λ = 6/R
λ = R/6
The Bohr model of atoms
Assumes that the angular momentum of electrons is quantized
Uses Einstein’s photoelectric equation
Predict continuous emission spectra for atoms
Predicts the same emission spectra for all types of atoms
The ground state energy H-atom is 13.6 eV. The energy needed to ionize H –atom from its second excited state is
12.1 eV
