Exams › NEET › Chemistry

The momentum of a particle having a de Broglie wavelength of 10⁻¹⁷ metres is [Given h = 6.625 × 10⁻³⁴ Js]:

1. 3.3125 × 10⁻¹⁷ kg ms⁻¹
2. 26.5 × 10⁻¹⁷ kg ms⁻¹
3. 6.625 × 10⁻¹⁷ kg ms⁻¹
4. 13.25 × 10⁻¹⁷ kg ms⁻¹

Correct answer: 6.625 × 10⁻¹⁷ kg ms⁻¹

Solution

The de Broglie wavelength (λ) is related to momentum (p) by the formula λ = h/p. Rearranging, p = h/λ. Substituting h = 6.625 × 10⁻³⁴ Js and λ = 10⁻¹⁷ m, we get p = (6.625 × 10⁻³⁴) / (10⁻¹⁷) = 6.625 × 10⁻¹⁷ kg ms⁻¹.

Related NEET Chemistry questions

• Protons and neutrons are also called
• From the \( \alpha \) -particle scattering experiment, Rutherford concluded that: This question has multiple correct options
• Which of following was not explained by Rutherford?
• The energy of an electron in the ground state of \( \boldsymbol{H} \) atom is \( -\mathbf{1 3 . 6 e V} \) The negative sign indicates that:
• Outer electronic configuration of Cl⁻ is:
• The sub-shell with lowest value of (n + l) is filled up first. When two or more sub-shells have the same (n + l) value, the sub-shell with lowest value of 'n' is filled up first. Therefore, the correct order is:

⚔️ Practice NEET Chemistry free + battle 1v1 →