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
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
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
Energy E of a hydrogen atom with principal quantum number n is given by .The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen,is approximately
1.5 eV
0. 85 eV
3.4 eV
1.9 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
6.8 eV
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
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
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
The ionization energy of hydrogen atom is 13.6 eV,What will be the ionization energy of helium atom ?
13.6 eV
27.2 eV
54.4 eV
The valence electron in alkali metal is a
f – electron
p – electron
s – electron
d – electron
