A proton and an α -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

Question

A proton and an α -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be (A) 1: 2 (B) 2 √2 : 1 (C) 2: 1 (D)  √2 : 1

Tardigrade Admin · Accepted Answer

Here, q V=(1/2) m v2 or m v=√2 q m V So, de-Broglie wavelength, λ=(h/m v)=(h/√2 q m V) i.e, λ propto (1/√q m) Hence, (λp/λa)=√(2 e × 4 m/e × m)=2 √2

Q. A proton and an $α$ -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

1 : 2

2 $2$ : 1

2 : 1

$2$ : 1

Here, $q V = \frac{1}{2} m v^{2}$
or $m v = 2 q mV$
So, de-Broglie wavelength,
$λ = \frac{h}{m v} = \frac{h}{2 q mV}$
i.e, $λ \propto \frac{1}{q m}$
Hence, $\frac{λ _{p}}{λ _{a}} = \frac{2 e \times 4 m}{e \times m} = 22$

Q. A proton and an α -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

1 : 2

2 2​ : 1

2 : 1

2​ : 1

Here, qV=21​mv2 or mv=2qmV​ So, de-Broglie wavelength, λ=mvh​=2qmV​h​ i.e, λ∝qm​1​ Hence, λa​λp​​=e×m2e×4m​​=22​

Q. A proton and an $α$ -particle are accelerated through same voltage. The ratio of their de - Broglie wavelength will be

2 $2$ : 1

$2$ : 1

Here, $q V = \frac{1}{2} m v^{2}$
or $m v = 2 q mV$
So, de-Broglie wavelength,
$λ = \frac{h}{m v} = \frac{h}{2 q mV}$
i.e, $λ \propto \frac{1}{q m}$
Hence, $\frac{λ _{p}}{λ _{a}} = \frac{2 e \times 4 m}{e \times m} = 22$