Question

The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong ?

Answer 1

1 eV = (1.6 x 10^-19 coulomb) x (1 volt)

= 1.6 x 10^-19 joule

15 keV = 18 x 10^-16 J …..(a)

Given that photon energy = hf = 15 keV

hf = hc/lambda …… (b)

Now hc = (6.63 x 10^-34 J-sec) x (3 x 10^8 m/sec)

= 19.89 x 10^-28 J-m …..(c)

From (b), lambda = hc/15 keV and from (a) & (c)

Lambda = (19.89 x 10^-28 J-m)/(18 x 10^-16 J)

= 1.105 x 10^-12 m

= 0.01105 Angstrom, which is in the x-ray range

Answer 2

Given:

Energy of em waves $\[ = 15\,keV\]$

To Find: Part of the spectrum to which the waves belong

Solution:

The waves produced when an electric field comes in contact with the magnetic field are known as electromagnetic waves.

The wavelength of an electromagnetic wave is given as:

$\[\lambda = \frac{{hc}}{E}\]$

The value of $hc$ in $eV$ is

Therefore, the wavelength of the given em wave can be calculated as:

$\[\begin{gathered} \lambda = \frac{{1240\,eVnm}}{{15 \times {{10}^3}\,eV}} \hfill \\ \lambda = 0.083\,nm \hfill \\ \end{gathered} \]$

The range of wavelengths of the different parts of the spectrum is given in the figure below.

As it is clearly visible that the value of $\[\lambda \]$ lies between $\[10\,nm\]$ and $\[0.001\,nm\]$, therefore, it belongs to the X-ray part of the spectrum.

Hence, the given em waves of energy of the order $\[15\,keV\]$ belong to the  X-ray part of the spectrum.

#SPJ2