Exams › JEE Main › Chemistry

A lithium ion and a proton are each accelerated through the same potential difference. If their de Broglie wavelengths are λ₁ and λ₂, respectively, and m_(Li) = 9mₚ, what is the ratio λ₁: λ₂?

1. 1: 2
2. 1: 4
3. 1: 1
4. 1: 3√(3)

Correct answer: 1: 3√(3)

Solution

The de Broglie wavelength is inversely proportional to the momentum of a particle, which depends on its mass and velocity. Since the lithium ion has a mass 9 times that of the proton and both are accelerated through the same potential difference, the proton will have a higher velocity and thus a shorter wavelength, leading to the ratio of their wavelengths being 1: 3√3.

Related JEE Main Chemistry questions

• At what wavenumber does the first line of the hydrogen Balmer series occur in the atomic spectrum?
• In the hydrogen emission spectrum, a spectral series has its limit at 12186.3 cm⁻¹. This series is identified as
• Two rapidly moving particles X and Y have de Broglie wavelengths of 1 nm and 4 nm, respectively. If the mass of X is nine times the mass of Y, what is the ratio of their kinetic energies, X: Y?
• Given that Planck’s constant is 6.63 × 10⁻³⁴ J s and the speed of light is 3.0 × 10⁸ m s⁻¹, which of the following is nearest to the wavelength, in nanometres, of light having frequency 8 × 10¹⁵ s⁻¹?
• The ionisation energy of He⁺ is 19.6 × 10⁻¹⁸ J atom⁻¹. What is the energy of the first stationary state (n = 1) of Li²⁺?
• What is the kinetic energy of an electron in the second Bohr orbit of a hydrogen atom, taking a₀ as the Bohr radius?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →