Atoms

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Question
CBSEENPH12039477

As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion:

  • kinetic energy, potential energy and total energy decrease

  • kinetic energy decreases, potential energy increases but total energy remains same

  • kinetic energy and total energy decrease but potential energy increases

  • its kinetic energy increases but the potential energy and total energy decrease

Solution

A.

kinetic energy, potential energy and total energy decrease

As we know that kinetic energy of an electron is 
KE ∝ (Z/n)
when the electron makes the transition from an excited state of the ground state then n, decreases and KE increases. We know that PE is lowest for the ground state. As TE=- KE and TE also decreases.

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Question
CBSEENPH12039545

Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is

  • 36.3 eV

  • 108.8 eV

  • 122.4 eV

  • 12.1 eV

Solution

B.

108.8 eV

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Question
CBSEENPH12039505

Hydrogen (1H1), deuterium (1H2), singly ionised helium (2He4+) and doubly ionised lithium (3Li8)2+ all have one electron around the nucleus. Consider an electron transition from n =2 to n=1. If the wavelengths of emitted radiation are λ123 andλ4, respectively for four elements, then approximately which one of the following is correct?

  •  4λ1=2λ2=2λ3 =λ4

  •  λ1=2λ2=2λ3 =λ4

  •  λ12=4λ3 =9λ4

  •  λ1=2λ2=3λ3 =4λ4

Solution

C.

 λ12=4λ3 =9λ4

For hydrogen atom, we get
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Question
CBSEENPH12039564

If a source of power 4kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called

  • X -rays

  • ultraviolet rays

  • microwaves

  • γ rays

Solution

A.

X -rays

As power of source = 4 x 103 = 1020 x hv
(∴ P/E = Number of photons)
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Question
CBSEENPH12039519

In a hydrogen like atom, electron makes the transition from an energy level with quantum number n to another with a quantum number (n – 1). If n >> 1, the frequency of radiation emitted is proportional to 

  • 1/n

  • 1/n2

  • 1/n3/2

  • 1/n3

Solution

D.

1/n3

An energy gap, ΔE = hv
Here, h is Planck's constant
therefore,
Frequency
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