The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is (1/√ m ), the value of m is

Question

The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is (1/√ m ), the value of m is (A) 4 (B) 16 (C) 8 (D) 2

Tardigrade Admin · Accepted Answer

(λα/λp)=(h/√(2mαqαV)(h)√2mpqpV (λα/λp)=√/(1)8) m =8

Q. The ratio of de-Broglie wavelength of an $α$ -particle and a proton accelerated from rest by the same potential is $\frac{1}{m}$ , the value of $m$ is

4

16

8

2

$\frac{λ _{α}}{λ _{p}} = \frac{h}{\frac{2 m _{α} q _{α} V}{\frac{h}{2 m _{p} q _{p} V}}}$
$\frac{λ _{α}}{λ _{p}} = \frac{1}{8} m = 8$

Q. The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is m​1​, the value of m is

4

16

8

2

λp​λα​​=2mp​qp​V​h​2mα​qα​V​​h​ λp​λα​​=81​​m=8

Q. The ratio of de-Broglie wavelength of an $α$ -particle and a proton accelerated from rest by the same potential is $\frac{1}{m}$ , the value of $m$ is

$\frac{λ _{α}}{λ _{p}} = \frac{h}{\frac{2 m _{α} q _{α} V}{\frac{h}{2 m _{p} q _{p} V}}}$
$\frac{λ _{α}}{λ _{p}} = \frac{1}{8} m = 8$