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
Bohr's model is based on :
Conservation of linear momentum
Conservation of angular momentum
Conservation of quantum frequency
Conservation of energy
Which of the following transitions gives photon of maximum energy ?
n = 1 to n = 2
n = 2 to n = 1
n = 2 to n = 6
n = 6 to n = 2
The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is
3.4 eV
6.8 eV
10.2 eV
Zero
The total energy of an electron in the first excited state of hydrogen is about – 3.4 eV. Its kinetic energy in this state is
-3.4 eV
- 6.8 e V
The ground state energy of H – atom is 13. 6 eV. The energy needed to ionize H – atom from its second excited state
1.51 eV
13.6 eV
12.1 eV
In terms of Bohr radius α0, the radius of the second Bohr orbit of a hydrogen atom is :
4 α0
8 α0
√ 2 α0
2 α0
The radius of hydrogen atom in its ground state is 5.3 x 10 -11 m. After collision with an electron it is found to have a radius of 21.2 x 10 – 11 m. What is the principal quantum number n of the final state of the atom ?
n = 4
n = 2
n = 16
n = 3
The ionization energy of the electron in the hydrogen atom in its ground state is 12.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
n = 3 to n = 2 states
n = 3 to n = 1 states
n = 2 to n = 1 states
n = 4 to n = 3 states
Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be :
2
3
4
1
