An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is
$2.5 A^{0}$
$0.25 n m$
$0.25 c m$
$0.025 n m$

Question

An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is
$2.5 A^{0}$
$0.25 n m$
$0.25 c m$
$0.025 n m$

Toppr Admin · Accepted Answer

-x3BB-cutoff-hcev-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-xA0-xA0-20-xD7-10-x2212-261-6-xD7-10-x2212-19-xD7-5-xD7-104-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-xA0-xA0-0-025nmSo- the answer is option -D-

An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is2.5A00.25nm0.25cm0.025nm

λcutoff=hcev =20×10−261.6×10−19×5×104 =0.025nmSo, the answer is option (D).

An electron beam in X-ray tube is accelerated through a potential difference of 50,000volt. These are then made to fall on a tungsten target. The shortest wavelength of the X-rays emitted by the tube is
$2.5 A^{0}$
$0.25 n m$
$0.25 c m$
$0.025 n m$

$λ_{c u t o f} f = \frac{h c}{e v}$

$= \frac{20 \times 10^{- 26}}{1.6 \times 10^{- 19} \times 5 \times 10^{4}}$

$= 0.025 n m$
So, the answer is option (D).