21311.The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1; hence the energy of fourth Bohr orbit would be
–41 kJ mol–1
–1312 kJ mol–1
–164 kJ mol–1
–82 kJ mol–1
21312.The de-Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 meters per second is approximately
10–33 m
10–31 m
10–16 m
10–25 m
21315.β emission is always accompanied 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
21316.(32Ge76, 34Se76) and (14Si30, 16S32) are examples of
isotopes and isobars
isobars and isotones
isotones and isotopes
isobars and isotopes
21317.What is the wavelength (in m) of a particle of mass 6.62 × 10–29 g moving with a velocity of 103 ms–1? (h = 6.62 × 10–34 Js)
6.62 ×10–4
6.62 ×10–3
10–5
105
21318.The emission spectrum of hydrogen discovered first and the region of the electromagnetic spectrum in which it belongs, respectively are
Lyman, ultraviolet
Lyman, visible
Balmer, ultraviolet
Balmer, visible
21319.As per de Broglie’s formula a macroscopic particle of mass 100 gm and moving at a velocity of 100 cm s–1 will have a wavelength of
6.6 × 10–29cm
6.6 × 10–30cm
6.6 × 10–31cm
6.6 × 10–32cm
21320.The statement that is NOT correct is
Angular quantum number signifies the shape of the orbital
Energies of stationary states in hydrogen like atoms is inversely proportional to the square of the principal quantum number
Total number of nodes for 3s orbital is three.
The radius of the first orbit of He+ is half that of the first orbit of hydrogen atom.