β- emission is always accompained by

• formation of antineutrino and α-particle

• emission of α-particle and γ-ray

• formation of antineutrino and γ-ray

• formation of antineutrino and positron

The value of Planck's constant is 6.63 x 10^-34 Js. The speed of light is 3 x 10¹⁷ nms^-1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 x 10¹⁵ s^-1?

• 10

• 25

• 50

• 75

What is the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m =-1.

• 10

• 6

• 4

• 2

An atomic nucleus having low n/p ratio tries to find stability by

• the emission of an α-particle

• the emission of a positron

• capturing an orbital electron (K-electron capture)

• emission of β -particle

(₃₂Ge⁷⁶, ₃₄Se⁷⁶) and (₁₄Si³⁰, ₁₆S³²) are examples of

• isotopes and isobars

• isobars and isotones

• isotones and isotopes

• isobars and isotopes

₉₈Cf²⁴⁶ was formed along with a neutron when an unknown radioactive substance was bomabrded with ₆C¹². The unknown substance is

• ₉₁Pa²³⁴

• ₉₀Th²³⁴

• ₉₂U²³⁵

• ₉₂U²³⁸

157.

Based on equation $\mathrm{E} = -2.178 \times {10}^{-18} \mathrm{J} \left(\frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\right)$ certain conclusions are written. Which of them is not correct?

• The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.

• Larger the value of n, the larger is the orbit radius

• Equation can be used to calculate the change in energy when the electron changes orbit.

• For n = 1 the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.

The energy of an electron in first Bohr orbit of H-atom is-13.6 eV. The possible energy value of electron in the excited state of Li²⁺ is:

• - 122.4 eV

• 30.6 eV

• - 30.6 eV

• 13.6 eV

The representation of the ground state electronic configuration of He by box-diagram as 

is wrong because it violates:

• Heisenberg's uncertainty principle

• Bohr's quantization theory of angular momenta

• Pauli exclusion principle

• Hund's rule

The electronic transitions from n = 2 to n = 1 will produce shortest wavelength in (where n = principal quantum state)

• Li²⁺

• He⁺

• H

• H⁺