Question

When $40kV$ is applied across a X-ray tube, X-ray is obtained with a maximum frequency of $9.7 \times {10^{18}}Hz$. Calculate the value of Plank’s constant from these data.

Answer 1

When a light containing some energy falls on the surface of an X-ray tube the phenomenon of photoelectric emission takes place. The formula for the energy of the photon that gets ejected when photoelectric effect occurs is applied in order to find the value of the Plank’s constant.

Formula used:
The energy of the photons is given by:
$E = hv$
Where, $h$ is the Plank’s constant and $v$ is the frequency of incident light.

Complete step by step answer:
The above problem revolves around the concept of photoelectric effect. In order to find the value for the Plank’s constant we first need to know the concept behind photoelectric emission of light. The phenomenon of emission of electrons from a metal surface when light is incident on it is called the photoelectric effect and this effect depends on the frequency of the incident light. The phenomenon of emission of electrons from a metal surface when electromagnetic radiations of sufficiently high frequency are incident on it is known as photoelectric effect. Photons or light generated electrons are emitted out once the light is incident on the surface.

The photoelectric effect takes place at only certain conditions. Different substances undergo photoelectric effect and emit photons only when exposed to radiations of frequency which is greater than the threshold frequency and only when it surpasses the minimum required energy needed to emit photons and this is given by the work function of the material. The work function of a metal is the minimum amount of energy required by an electron to just escape from its surface.

Since this energy is dependent on the frequency of the incident ray of light an equation is obtained relating them. The equation for the energy of the emitted photons is given by:
$E = hv$ --------($1$)
Here, in this question when light with frequency greater than the threshold frequency is assumed to hit the surface of an X-ray tube and thus photoelectric emission takes place where the ejected photons will contain an energy given by the equation above, that is, equation ($1$)

Let us now extract the data given in the question. The energy which is applied on the X-ray tube and frequency of the incident energy with which the X-ray is obtained is given. We are asked to find the Plank’s constant for the material. Given, $E = 40kV$ and $v = 9.7 \times {10^{18}}Hz$. By rearranging the terms of equation ($1$) to make $h$ as the subject we get:
$h = \dfrac{E}{v}$ ------($2$)
By substituting the given values in equation ($2$) we get:
$h = \dfrac{{40 \times {{10}^3}}}{{9.7 \times {{10}^{18}}}}$
We solve the equation to get the value for the Plank’s constant.
$\Rightarrow h = \dfrac{{40}}{{9.7 \times {{10}^{18 - 3}}}}$
$\Rightarrow h = \dfrac{{40}}{{9.7 \times {{10}^{15}}}}$
$\Rightarrow h = 4.123 \times {10^{ - 15}}Js$
$\therefore h \approx 4.12 \times {10^{ - 15}}Js$

Hence the Plank’s constant for the material on which photoelectric effect takes place through, that is, the X-ray tube is given by the value $4.12 \times {10^{ - 15}}\,Js$.

Note: A common confusion that occurs in this problem is the consideration of the standard or the constant value for Plank’s constant which is $6.626 \times {10^{ - 34}}Js$.This value is not required to be applied here because even-though this value is mostly considered for many materials obeying photoelectric effect, there are certain materials which have a unique or a different value of Plank’s constant which is dependent on the energy given to the substance and its corresponding frequency.