If the energy of an X-ray photon is $"100 keV"$, then if its wavelength matches that of the ejected core electron, what should be the kinetic energy of that electron?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

About $"9.78 keV"$, assuming $100%$ transfer of energy.

The de Broglie wavelength $lambda$ is given by

$lambda = h/p = h/(mv)$,

where:

  • $h = 6.626 xx 10^(-34) "J"cdot"s"$ is .
  • $p = mv$ is the linear momentum in $"kg"cdot"m/s"$, for mass $m$ and velocity $v$.

Given the energy of an X-ray photon as $"100 keV"$, we have:

$E_"photon" = "100 keV" = hnu = (hc)/lambda$

where $nu$ is the frequency of the photon in $"s"^(-1)$ and $c = 2.998 xx 10^(8) "m/s"$ is the speed of light.

And so, we suppose that the photon wavelength is numerically the same as the electron wavelength. I suppose this could occur if one tried to ionize a core electron for X-ray diffraction.

$lambda = (hc)/E_"photon" = h/(p)$

This gives a wavelength of:

$lambda = (6.626 xx 10^(-34) cancel"J"cdotcancel"s" cdot 2.998 xx 10^(8) "m/"cancel"s")/("100 "cancel"k"cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV") xx (1000 cancel"J")/cancel"1 kJ")$

$= 1.240 xx 10^(-11) "m"$

Knowing the wavelength, we consider the kinetic energy expressed as a function of linear momentum:

$K = 1/2 mv^2 = p^2/(2m)$

Thus, with $p = h/lambda$ from the de Broglie relation, and an electron rest mass of $9.109 xx 10^(-31) "kg"$, we have that:

$color(blue)(K) = (h//lambda)^2/(2m)$

$= ((6.626 xx 10^(-34) "kg"cdot"m"^(cancel(2))"/s")/(1.240 xx 10^(-11) cancel"m"))^2/(2 cdot 9.109 xx 10^(-31) cancel"kg")$

$=$ $color(blue)ul(1.567 xx 10^(-15) "J")$

Or perhaps in more useful units...

$1.567 xx 10^(-15) cancel"J" xx (cancel"1 eV")/(1.602 xx 10^(-19) cancel"J") xx ("1 keV")/(1000 cancel"eV")$

$=$ $color(blue)ul("9.782 keV")$

So the X-ray electron has a kinetic energy less than $1//10$th of the energy of an X-ray photon, assuming $100%$ transfer of energy.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
A photon of light has a wavelength of 0.050 cm. What is its energy?

First find the frequency of the photon from the equation $c=flambda$ $therefore f=c/lambda=(3xx10^8m//s)/(0.050xx10^(-2)m)=6xx10^11Hz$. Energy of such photons is quantized and given by $E=hf$ $=(6.6xx10^(-34))(6xx10^11)$ $=3.96xx10^(-22)J$

1 Answers 1 views
Microwave ovens emit microwave energy with a wavelength of 12.2 cm. What is the energy of exactly one photon of this microwave radiation?

Let us calculate the frequency $nu$ of the photon. We have the expression between the velocity of light in vacuum $c=nuxxlambda$ or $nu=c/lambda$......(1) where $lambda$ is the wave-length of...

1 Answers 1 views
An emission ray has a frequency of $5.10 * 10^14$ $Hz$. Given that the speed of light is $2.998 * 10^8$ $m$$/s$, what is the wavelength of the ray?

You're given a frequency $(f)$ of $5.10 * 10^(14) Hz$, and you're given a speed $(v)$ of $2.998 * 10^8$ $m/s$ Plug the data into the equation to solve...

1 Answers 1 views
Consider an electromagnetic wave with a frequency of $3.0xx10^10$ $Hz$. What is its wavelength? What is the energy of one photon of this light?

For electromagnetic radiation: $c=flambda$ where $c$ is the speed of light, $f$ is the frequency, $lambda$ is the wavelength. Rearranging: $lambda = c/f=(3.0xx10^8)/(3.0xx10^10)=10^-2$ $m$ The energy of a photon is...

1 Answers 1 views
The work function for mercury is equivalent to 435kJ/mol photons. What is the kinetic energy, in joules, of the ejected electrons when light of 188 nm strikes a mercury surface?

Einstein photo electric equation $E_k+w=hf$ Where $E_k->"KE of ejected electron"$ $w->"photo electric work function"=435" kJ/mol"$ $=(435xx1000J/"mol")/(6.023xx10^23mol^-1)=7.22xx10^-19J$ $h->"Planck's constant"=6.62xx10^-34Js$ $f->"frequency of incident light"=c/lambda$ $=(3xx10^8m/s)/(188xx10^-9m)=3/188xx10^17Hz$ where velocity...

1 Answers 1 views
An electron in an hydrogen atom makes a transition from the ground state, $n=1$, to the excited state, $n=5$. Calculate the energy in Joules, frequency in Hertz $(1/s)$, and wavelength in $nm$ of the photon?

Wavelength You use the Rydberg formula to calculate the wavelength, λ: $color(blue)(bar(ul(|color(white)(a/a) 1/λ = RZ^2(1/n_2^2 -1/n_1^2)color(white)(a/a)|)))" "$ where $R =$ the Rydberg constant ($1.097 × 10^7color(white)(l)...

1 Answers 1 views
What is the kinetic energy for one electron whose wavelength is $"650 nm"$? Its rest mass is $9.109 xx 10^(-31) "kg"$.

The De Broglie equation states that the $"KE" = (1/2)"mv"^2$ $"Where KE is kinetic energy"$ $"Where m is mass and is a constant for each particle"$ $"Where v is...

1 Answers 1 views
A photon on a hydrogen atom excited an electron from $n = 1$ to make a transition to $n=3$. What is the wavelength (in run) pertaining to this transition and was the transition emitting or absorbing in nature?

In order to solve this question we will use the Bohr Model. The change in energy according to Bohr can be calculated by: $DeltaE=-2.178xx10^(-18)Z^(2)J(1/(n_("final")^2)-1/(n_("initial")^2))$ Here, the initial level is $n_("initial")=1$...

1 Answers 1 views
In a sample of excited hydrogen atoms electrons make transition from n = 3 to n = 1 and emitted photons strike on a metal of work function 6.08 eV. The maximum kinetic energy of the ejected electron in the process will be?

The idea here is that in order for an electron to be ejected from the surface of the metal, the energy of the incoming photon must exceed the work function...

1 Answers 1 views
An electron in a mercury atom jumps from level $a$ to level $g$ by absorbing a single photon. How do you determine the energy of the photon in joules?

Relativistic effects are present here... but I assume that the transition from level $a$ to level $g$: has $DeltaL = pm 1$ (the total change in angular momentum is...

1 Answers 1 views
X