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NEET
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NEET
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Physics
Chapter
Atomic Physics

NEET · Physics · Atomic Physics

Given below are two statements Statement I: In hydrogen atom, the...

Given below are two statements

Statement I: In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1)\left({E}_{1}\right) to higher energy orbit \left({E}_{2}\right),isgivenas, is given ashf = {E}_{1} - {E}_{2}</p><p><strong>StatementII:</strong>Thejumpingofelectronfromhigherenergyorbit</p><p><strong>Statement II: </strong>The jumping of electron from higher energy orbit\left({E}_{2}\right)tolowerenergyorbit(E1)to lower energy orbit \left({E}_{1}\right) is associated with frequency of radiation given as f=(E2−E1)hf = \dfrac{\left({E}_{2} - {E}_{1}\right)}{h}. This condition is Bohr's frequency condition.

In the light of the above statements, choose the correct answer from the options given below:

Given below are two statements Statement I: In hydrogen atom, the frequency of radiation emitted when an electron jumps

Options

  1. ABoth statement I and statement II are true.
  2. BBoth statement I and statement II are false.
  3. CStatement I is correct but statement II is false.
  4. DStatement I is incorrect but statement II is true.

Correct answer

D. Statement I is incorrect but statement II is true.

Step-by-step solution

Given below are two statements Statement I: In hydrogen atom, the frequency of radiation emitted when an electron jumps 

According to Bohr's model, when electron makes the transition from lower orbit to higher orbit, its energy can be given as

 ∆E=Efinal−Einitial=hf∆ E = {E}_{final} - {E}_{initial} = hf, here, hh is Planck's constant.

Then, f=E2−E1hf = \dfrac{{E}_{2} - {E}_{1}}{h}.

Bohr's frequency condition is the law that state that the frequency of the radiation emitted or absorbed during the transition of an atomic system between two stationary states equals the difference in the energies of the states divided by Planck's constant.

Hence, Statement I is incorrect but statement II is true.

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