Atoms · Question 3

Angular momentum of an electron in a hydrogen atom is 3h , then the energy of the electron is eV.

1. -1.51
2. -0.85
3. -0.38
4. -0.28

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom .

Two electrons are moving in orbits of two hydrogen-like atoms with speeds 3 105 m/s and 2.5 105 m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are respectively.

1. 8 and 10
2. 10 and 12
3. 9 and 8
4. 6 and 5

In hydrogen atom spectrum, (R Rydberg's constant) Choose the correct answer from the options given below :

1. A, B Only
2. B, D Only
3. A, B and D Only
4. A, B and C Only

A 7.9 MeV -particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) m. [ 14 0 = 9 109 N m2/C2 . and electron charge . = 1.6 10-19 C ]

1. 2.88 10-14
2. 5.76 10-14
3. 1.44 10-13
4. 1.69 10-12

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly nm.

1. 1784
2. 1875
3. 1217
4. 1550

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is m. (Atomic number of gold = 79 and 14 0 = 9 109 in SI units)

1. 2.95 10-16
2. 3.85 10-14
3. 2.95 10-14
4. 3.85 10-16

The energy of an electron in an orbit of the Bohr's atom is -0.04E0 eV where E0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2 Lh is :

1. 6
2. 2
3. 5
4. 4

In a hydrogen like ion, the energy difference between the 2nd excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

1. 1
2. 4
3. 3
4. 2

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

1. 5 : 27
2. 27 : 5
3. 3 : 4
4. 5 : 36