The energy that should be added to an electro | vedclass

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

Question

(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.From this relatio

Vedclass · Accepted Answer

Thrice the initial energy

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.From this relation,we have $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}}$.Given $\lambda = 1 \ nm$ and $\lambda' = 0.5 \ nm$,we substitute these values:$\frac{0.5}{1} = \sqrt{\frac{E}{E'}}$Squaring both sides,we get $\frac{1}{4} = \frac{E}{E'}$,which means $E' = 4E$.The energy that should be added is $\Delta E = E' - E = 4E - E = 3E$.Therefore,the energy to be added is thrice the initial energy.

The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

Similar Questions

Assume each reaction is carried out in an open container. For which reaction will $\Delta H = \Delta E$?

If an electric heater is rated at $1000\, W$,then the time required to heat one litre of water from $20\, ^oC$ to $60\, ^oC$ is

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. In order for a ray to suffer minimum deviation,it should be incident at an angle of.......$^{\circ}$.

What is obtained by the electrolysis of carnallite?

$ f: R \rightarrow R $ and $ g:[0, \infty) \rightarrow R $ are defined by $ f(x)=x^{2} $ and $ g(x)=\sqrt{x} $. Which one of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module