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Question

(A) The energy gained by an electron accelerated through a potential difference $V$ is given by $E = eV$.
In an $X-ray$ tube, this energy is converte

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$10^{19} \ Hz$

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(A) The energy gained by an electron accelerated through a potential difference $V$ is given by $E = eV$.
In an $X-ray$ tube, this energy is converted into the maximum energy of the emitted photon, which is $E = h
u_{max}$.
Equating the two, we get $h
u_{max} = eV$.
Therefore, the maximum frequency $
u_{max}$ is given by $
u_{max} = \frac{eV}{h}$.
Substituting the given values:
$
u_{max} = \frac{1.6 	imes 10^{-19} \ J 	imes 42,000 \ V}{6.63 	imes 10^{-34} \ J \cdot s}$.
$
u_{max} \approx \frac{6.72 	imes 10^{-15}}{6.63 	imes 10^{-34}} \approx 1.01 	imes 10^{19} \ Hz$.
Rounding to the nearest order of magnitude, the maximum frequency is $10^{19} \ Hz$.

$A$ potential difference of $42,000 \ V$ is used in an $X-ray$ tube to accelerate electrons. The maximum frequency of the $X-radiations$ produced is ($1 \ eV = 1.6 \times 10^{-19} \ J$ and $h = 6.63 \times 10^{-34} \ J \cdot s$).

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