Two particles A and B are in motion. If the w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the de Broglie equation,$\lambda = h / p$,where $\lambda$ is the wavelength,$h$ is Planck's constant,and $p$ is the momentum.Given $\

Vedclass · Accepted Answer

$1000 \ \mathring{A}$

Vedclass · Answer

(A) According to the de Broglie equation,$\lambda = h / p$,where $\lambda$ is the wavelength,$h$ is Planck's constant,and $p$ is the momentum.Given $\lambda_{A} = 5 	imes 10^{-8} \ m$ and $p_{B} = 0.5 \ p_{A}$.Since $\lambda \propto 1 / p$,we have $\lambda_{B} / \lambda_{A} = p_{A} / p_{B}$.Substituting $p_{B} = 0.5 \ p_{A}$,we get $\lambda_{B} / \lambda_{A} = p_{A} / (0.5 \ p_{A}) = 2$.Therefore,$\lambda_{B} = 2 	imes \lambda_{A} = 2 	imes 5 	imes 10^{-8} \ m = 10^{-7} \ m$.Converting to $\mathring{A}$: $1 \ m = 10^{10} \ \mathring{A}$,so $10^{-7} \ m = 10^{-7} 	imes 10^{10} \ \mathring{A} = 1000 \ \mathring{A}$.

Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5 \times 10^{-8} \ m$,calculate the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.

Similar Questions

The Davisson and Germer experiment showed that:

If the kinetic energy of an electron is $1.25 \times 10^{-28} \ J$,then what will be the de-Broglie wavelength of the electron? (Mass of electron $= 9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ Js$)

What should be the ratio of velocities of $CH_4$ and $O_2$ molecules so that they are associated with de Broglie waves of equal wavelength?

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

If $a_0$ is the Bohr radius,then the de-Broglie wavelength of an electron revolving in the second excited state of the $H$ atom will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module