Find the (a) maximum frequency (b) minimum wavelength of X-rays produced by 30 kV electrons.

Question Id: PH0000012

Question: Find the 

(a) maximum frequency

(b) minimum wavelength of X-rays produced by 30 kV electrons.

Answer:

(a) Maximum frequency of the X-ray produced by 30kV electron is $\ 7.253 \times 10^{18}Hz$.

(b) Minumum wavelength ( $\lambda$ ) is $0.41$ angstroms.

Explanation:

Given, potential $V=30kV$

                   $\Rightarrow V=30 \times 10^3V$

                   $\Rightarrow V=3 \times 10^4V$

So, energy of the X-ray is $E=3 \times 10^4eV$

[ $1eV=1.602\times 10^{-19}J$ ]

We know a particle having frequency ( $\nu$ ) contain energy $E=h\nu$ ; where h is Planck's constant $h=6.626\times 10^{-34}Js$

$$\Rightarrow \nu =\frac{E}{h}$$

$$\Rightarrow \nu =\frac{1.602\times 10^{-19} \times 3 \times 10^{4}}{6.626\times 10^{-34}}$$

$$\Rightarrow \nu =\frac{1.602 \times 3\times 10^{19} }{6.626}$$

$$\Rightarrow \nu =0.7253 \times 10^{19}$$

$$\Rightarrow \nu =7.253 \times 10^{18}$$

Hence, maximum frequency of the X-ray produced by 30kV electron is $\ 7.253 \times 10^{18}Hz$.

(b) We know the relationship between wavelength ( $\lambda$ ) and frequency ( $\nu$ ) is $c=\nu \lambda$ ; where c is velocity of light.

$$\Rightarrow \lambda =\frac{c}{\nu }$$

When frequency ( $\nu$ )is maximum then wavelength ( $\lambda$ ) is minimum

$$\Rightarrow \lambda_{min} =\frac{c}{\nu_{max} }$$

$$\Rightarrow \lambda_{min} =\frac{3\times 10^8}{7.253\times 10^{18} }$$

$$\Rightarrow \lambda_{min} =0.41\times 10^{-10}$$

$\Rightarrow \lambda_{min} =0.41$ angstroms

So, minumum wavelength ( $\lambda$ ) is $0.41$ angstroms.

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