The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1: √2. Then, the ratio of Vp to Vd will be

Question

The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1: √2. Then, the ratio of Vp to Vd will be (A) 1: 1 (B) √2: 1 (C) 2: 1 (D) 4: 1

Tardigrade Admin · Accepted Answer

Kinetic energy gained by a charged particle accelerated by a potential

V

is

q V

K E = q V

\Rightarrow \frac{p ^{2}}{2 m} = q V \Rightarrow p = 2 m q V

p = \frac{h}{λ}, thus λ = \frac{h}{2 m q V}

now

\frac{λ _{p}}{λ _{d}} = \frac{m _{d} V _{d}}{m _{p} V _{p}}

\Rightarrow \frac{1}{2} = \frac{2 V _{d}}{V _{p}} \Rightarrow \frac{V _{p}}{V _{d}} = 4

Q. The ratio of wavelengths of proton and deuteron accelerated by potential $V_{p}$ and $V_{d}$ is $1 : 2$ . Then, the ratio of $V_{p}$ to $V_{d}$ will be

$1 : 1$

$2 : 1$

$2 : 1$

$4 : 1$

Q. The ratio of wavelengths of proton and deuteron accelerated by potential Vp​ and Vd​ is 1:2​. Then, the ratio of Vp​ to Vd​ will be

1:1

2​:1

2:1

4:1

Kinetic energy gained by a charged particle accelerated by a potential V is qV KE=qV ⇒2mp2​=qV⇒p=2mqV​ p=λh​, thus λ=2mqV​h​ now λd​λp​​=mp​Vp​md​Vd​​​ ⇒2​1​=Vp​2Vd​​​⇒Vd​Vp​​=4

Q. The ratio of wavelengths of proton and deuteron accelerated by potential $V_{p}$ and $V_{d}$ is $1 : 2$ . Then, the ratio of $V_{p}$ to $V_{d}$ will be

$1 : 1$

$2 : 1$

$2 : 1$

$4 : 1$