\(λ_e > λ_p > λ_α\)
\(λ_α > λ_e > λ_p\)
\(λ_p = λ_α = λ_e\)
\(λ_p > λ_e > λ_α\)
The matter waves, the wavelength is associated with the microscopic particles like protons, electrons, neutrons, \(\alpha\)-particle etc., is or the order of \(10^{-10}m\).
The relation between de-Broglie wavelength \(\lambda\) and the kinetic energy \(K\) of the particle is given by:
\(λ = \frac{h}{m.v} = \frac{h}{√(2.m.K.E)}\)
\(\text{as K.E. is same } λ∝\frac{1}{\sqrt{m}}\)
mass of electron = \(9.1 × 10^{-31}\) kg
mass of proton = \(1.67 × 10^{-27}\) kg
mass of α-particle = \(6.68 × 10^{-27}\) kg
\(λ_e > λ_p > λ_α\)
So, the correct option is (A): \(λ_e > λ_p > λ_α\)
An alpha particle moves along a circular path of radius 0.5 mm in a magnetic field of \( 2 \times 10^{-2} \, \text{T} \). The de Broglie wavelength associated with the alpha particle is nearly
(Planck’s constant \( h = 6.63 \times 10^{-34} \, \text{Js} \))
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: