Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. If the circumference of orbit can be expressed in terms of de-Broglie wavelength λ ,then find out it's value ?
( 0.529 ) n λ
√ n λ
( 13.6 ) λ
n λ
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
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
3.4 eV
13.6 eV
12.1 eV
Energy levels A, B, C, of a certain atom correspond to increasing values of energy ie, EA < EB < EC, If λ1, λ2, λ3 are the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relation is correct ?
λ3 = λ1 +λ2
λ3 = λ1 λ2/λ1 +λ2
λ1 +λ2 + λ3 = 0
λ23 = λ21 +λ22
When a hydrogen atom is raised from the ground state to an excited state.
PE decreases and KE increases
PE increases and KE decreases
Both KE and PE decreases
Absorption spectrum
The ionization energy of hydrogen atom is 13.6 eV,What will be the ionization energy of helium atom ?
27.2 eV
6.8 eV
54.4 eV
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
Which source is associated with a line emission spectrum ?
Electric fire
Neon street sign
Red traffic light
Sun
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
The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is
10.2 eV
Zero
