The binding energy of the innermost electron in tungsten is 40 keV. To produce characteristic X-rays using a tungsten target in X-ray tube the potential difference 'V' between the cathode and the anti-cathode will be

V < 40 K V

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V > 40 K V

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V = 40 K V

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V = \sqrt{40} K V

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Solution

The correct option is B $V > 40 K V$
To knock out electron from innermost shell, the kinetic energy of hitting electron should be greater than the energy of binding of electron inner shell.
$E > 40 k e V$
$e V > 40 k e V$
$V > 40 k V$

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The binding energy of the innermost electron in tungsten is 40 keV. To produce characteristic X-rays using a tungsten target in X-ray tube the potential difference 'V' between the cathode and the anti-cathode will be

The correct option is B V>40KVTo knock out electron from innermost shell, the kinetic energy of hitting electron should be greater than the energy of binding of electron inner shell.E>40keVeV>40keVV>40kV

The correct option is B $V > 40 K V$
To knock out electron from innermost shell, the kinetic energy of hitting electron should be greater than the energy of binding of electron inner shell.
$E > 40 k e V$
$e V > 40 k e V$
$V > 40 k V$