JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 6)
An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
10$$-$$2 nm
10$$-$$1 nm
10$$-$$3 nm
10$$-$$4 nm
Explanation
Given, V = 1.24 million volt = 1.24 $$\times$$ 106 volt
Since, energy (E) = eV
where, e is the charge of electron = 1.6 $$\times$$ 10$$-$$19 C
$$\therefore$$ E = 1.6 $$\times$$ 10$$-$$19 $$\times$$ 1.24 $$\times$$ 106 ..... (i)
As we know that,
Energy of photon, $$E = {{hc} \over \lambda }$$ .... (ii)
Here, Planck's constant, h = 6.67 $$\times$$ 10$$-$$34 J-s,
c = speed of light in free space, c = 3 $$\times$$ 108 ms$$-$$1
Equating Eqs. (i) and (ii), we get
$$1.6 \times {10^{ - 19}} \times 1.24 \times {10^6} = {{6.67 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over \lambda }$$
$$ \Rightarrow \lambda = {{20.01 \times {{10}^{ - 13}}} \over {1.6 \times 1.24}} = 10.09 \times {10^{ - 13}}$$
$$ = 1.009 \times {10^{ - 12}} \simeq {10^{ - 3}} \times {10^{ - 9}}$$
= 10$$-$$3 nm
Since, energy (E) = eV
where, e is the charge of electron = 1.6 $$\times$$ 10$$-$$19 C
$$\therefore$$ E = 1.6 $$\times$$ 10$$-$$19 $$\times$$ 1.24 $$\times$$ 106 ..... (i)
As we know that,
Energy of photon, $$E = {{hc} \over \lambda }$$ .... (ii)
Here, Planck's constant, h = 6.67 $$\times$$ 10$$-$$34 J-s,
c = speed of light in free space, c = 3 $$\times$$ 108 ms$$-$$1
Equating Eqs. (i) and (ii), we get
$$1.6 \times {10^{ - 19}} \times 1.24 \times {10^6} = {{6.67 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over \lambda }$$
$$ \Rightarrow \lambda = {{20.01 \times {{10}^{ - 13}}} \over {1.6 \times 1.24}} = 10.09 \times {10^{ - 13}}$$
$$ = 1.009 \times {10^{ - 12}} \simeq {10^{ - 3}} \times {10^{ - 9}}$$
= 10$$-$$3 nm
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