If the momentum of an electron is changed by | vedclass

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

Question

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}.$Given that the wavelength changes by $0.50\%,$ the new wavelength $\lambda'$ is $\la

Vedclass · Accepted Answer

$199 \Delta p$

Vedclass · Answer

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}.$Given that the wavelength changes by $0.50\%,$ the new wavelength $\lambda'$ is $\lambda' = \lambda - \frac{0.5}{100} \lambda = 0.995 \lambda = \frac{199}{200} \lambda.$Since $\lambda = \frac{h}{p},$ the new momentum $p'$ is $p' = p + \Delta p.$Thus,$\lambda' = \frac{h}{p + \Delta p}.$Equating the two expressions for $\lambda':$$\frac{h}{p + \Delta p} = \frac{199}{200} \left( \frac{h}{p} \right).$$\Rightarrow \frac{1}{p + \Delta p} = \frac{199}{200p}.$$\Rightarrow 200p = 199(p + \Delta p).$$\Rightarrow 200p = 199p + 199 \Delta p.$$\Rightarrow p = 199 \Delta p.$

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

Similar Questions

If the kinetic energy of a free electron is doubled,by what factor does its de Broglie wavelength change?

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

What is the kinetic energy of a neutron associated with a de Broglie wavelength of $1.40 \times 10^{-10} \ m$? (Mass of neutron $m = 1.675 \times 10^{-27} \ kg$,$h = 6.6 \times 10^{-34} \ J \cdot s$).

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

De Broglie waves are associated with moving particles. These particles can be .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module