The ratio of de-broglie wavelength of a deuterium atom to that of a-particle when the velocity of former is five times greater than that of latter is:

Question

The ratio of de-broglie wavelength of a deuterium atom to that of a-particle when the velocity of former is five times greater than that of latter is: (A) 4 (B) 10.2 (C) 2 (D) 0

Tardigrade Admin · Accepted Answer

λ=( h / mv ) mass of α -particle =4u, mass of deuterium =2u V D =5 Vα λ D =(1/2 × V D ), λα=(1/4 V α) (λ D /λα)=(4 V α/2 V D ) =(4/2 × 5)=(4/10)=0.4

Q. The ratio of de-broglie wavelength of a deuterium atom to that of a-particle when the velocity of former is five times greater than that of latter is:

$4$

$10.2$

$2$

$0$

$λ = \frac{h}{m v}$

mass of $α$ -particle $= 4 u$ , mass of deuterium $= 2 u$

$V_{D} = 5 V_{α}$

$λ_{D} = \frac{1}{2 \times V _{D}}, λ_{α} = \frac{1}{4 V _{α}}$

$\frac{λ _{D}}{λ _{α}} = \frac{4 V _{α}}{2 V _{D}}$

$= \frac{4}{2 \times 5} = \frac{4}{10} = 0.4$

Q. The ratio of de-broglie wavelength of a deuterium atom to that of a-particle when the velocity of former is five times greater than that of latter is:

4

10.2

2

0

λ=mvh​ mass of α -particle =4u, mass of deuterium =2u VD​=5Vα​ λD​=2×VD​1​,λα​=4Vα​1​ λα​λD​​=2VD​4Vα​​ =2×54​=104​=0.4

$4$

$10.2$

$2$

$0$

$λ = \frac{h}{m v}$

mass of $α$ -particle $= 4 u$ , mass of deuterium $= 2 u$

$V_{D} = 5 V_{α}$

$λ_{D} = \frac{1}{2 \times V _{D}}, λ_{α} = \frac{1}{4 V _{α}}$

$\frac{λ _{D}}{λ _{α}} = \frac{4 V _{α}}{2 V _{D}}$

$= \frac{4}{2 \times 5} = \frac{4}{10} = 0.4$