Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.

Physics

1 answer:

ivolga24 [154]3 years ago

8 0

Answer: 2.068* $10^{-14}$ m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602* $10^{-16}$

h= planck constant = 6.626* $10^{-34}$

c= speed of light =2.998* $10^{8}$

v= potential difference= 6* $10^{4}$

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= $\frac{hc}{qv}$

λ = 6.626* $10^{-34}$ * 2.998* $10^{8}$ / 1.602* $10^{-16}$ * 6* $10^{4}$

λ = $\frac{19.878*10^{-26} }{9.612*10^{-12} }$

by doing the necessary calculations, we have that

λ = 2.068* $10^{-14}$ m

You might be interested in

Select the correct answer. Which healthy snack can provide protein after physical activity? A. an orange slice B. pretzels C. yo

givi [52]

Answer:

From the given choices, the healthy snack that can provide protein after physical activity is the third option, yogurt. Yogurts are derived from fermentation of raw materials by the healthy bacteria, usually of lactobacillus origin. Other items in the choices are not able to provide protein.

Explanation:

4 0

3 years ago

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$