Question

If kinetic energy of an electron is increased by 44% find the percentage change in the broccoli constant or wavelength

Answer 1

Answer:

the correct answer is 16.6

Answer 2

Given:

The kinetic energy of an electron is increased by 44%.

To Find:

The percentage change in the broccoli constant or wavelength.

Solution:

To find the we will follow the following steps:

As we know,

Kinetic energy and momentum relation are given by:

$Kinetic \: energy= \frac{ {p}^{2} }{2m}$

Debriglies wavelength and momentum relationship =

$lambda( l) = \frac{h}{p}$

$lambda( l) = \frac{h}{ \sqrt{2mk} }$

Here,

p = momentum

lambda(l) = wavelength

k = kinetic energy

m = mass of the object

h =

Final kinetic energy k2=

$\frac{40}{100}k1 + k1$

Final kinetic energy k2 = 1.44k1

wavelength inversely proportional to kinetic energy so,

$\frac{l1}{l2} = \frac{k2}{k1}$

Now, putting values in the above equation we get,

$\frac{l1}{l2} = \frac{1.44k1}{k1}$

$\frac{l2}{l1} = \frac{100}{144}$

$1 - \frac{l2}{l1} = 1 - \frac{100}{144}$

$\frac{l1 - l2}{l1} = \frac{144 - 100}{144}$

$\frac{l1 - l2}{l1} = \frac{44}{144}$

The percentage change in de Broglie's wavelength is

$\frac{44}{144} \times 100 =30.5%$

Henceforth, the percentage change in the broccoli constant or wavelength is 30.5%.