Question

Atomic number of anticathode material in an X−ray tube is $41$ . Wavelength of ${K_\alpha }$ X−ray produced in the tube is:
A) $0.66\mathop {\rm A}\limits^o$
B) $0.76\mathop {\rm A}\limits^o$
C) $0.82\mathop {\rm A}\limits^o$
D) $0.88\mathop {\rm A}\limits^o$

Answer 1

You can easily solve this question by trying to recall the equation of wavelength with Rydberg Constant and atomic number. This equation is a very useful and frequently occurring equation which you are supposed to learn by heart. Let us tell you the equation in case you don’t know:
$\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$

Complete step by step answer:
We will be trying to solve the question exactly like we told you in the hint section of the solution to this question. Firstly, we will have a look at what is given to us by the question and how we can use that given information to reach the answer. Then, we will use the corresponding equation to reach the answer to the question asked.
Now, let’s have a look at what is given to us in the question:
Atomic number $\left( z \right) = 41$
The question is asking about ${K_\alpha }$ X-ray, which means, the value of the shell will be ${n_2} = 2$ and it’ll fall into the shell ${n_1} = 1$ .
Having a look at the given information, we can say that we need to use the following formula to reach at the answer to this question:
$\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
We already know that $R$ is Rydberg Constant and its value is: $R = 1.097 \times {10^7}\,{m^{ - 1}}$
Now, we have all the information and knowledge that we need to solve this question, the only thing remaining is to use the formula and substitute the known values in it. So, let’s do it:
$\dfrac{1}{\lambda } = R{\left( {z - 1} \right)^2}\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]$
We already know that $z = 41$
We figured it out that ${n_1} = 1$ and ${n_2} = 2$
And we know the value of the Rydberg constant. Substituting the values, we get:
$\dfrac{1}{\lambda } = 1.097 \times {10^7}\left( {41 - 1} \right)\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right]\,{m^{ - 1}}$
Solving further, we get:
$\dfrac{1}{\lambda } = \dfrac{1}{{0.76}} \times {10^{10}}\,{m^{ - 1}}$
After reciprocating, we get:
$\lambda = 0.76 \times {10^{ - 10}}m = 0.76\mathop {\rm A}\limits^o$
Here, we can see that our answer matches the option B, hence, the correct option is option (B).

Note: The main thing that is tough in this question is figuring out the values of ${n_1}$ and ${n_2}$ , which can be processed from the fact that question has asked us about ${K_\alpha }$ X-ray. Other than that, you should always remember this formula as it is very frequently occurring in various exams. And, do not perform any calculations mistakes, since such questions are calculation heavy.