The maximum frequency of X-rays produced in an X-ray tube at an operating voltage of 25kV is
(h $= 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8}$ m/s)

$4.06 \times 10^{18} H z$
$9.2 \times 10^{18} H z$
$6.08 \times 10^{18} H z$
$3.02 \times 10^{18} H z$

Question

The maximum frequency of X-rays produced in an X-ray tube at an operating voltage of 25kV is
(h $= 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8}$ m/s)

$4.06 \times 10^{18} H z$
$9.2 \times 10^{18} H z$
$6.08 \times 10^{18} H z$
$3.02 \times 10^{18} H z$

Toppr Admin · Accepted Answer

Operating-xA0-voltage-xA0-h-x3BD-eoperating-xA0-voltage-xD7-eh-x3BD-x3BD-25-xD7-103-xD7-1-6-xD7-10-x2212-196-63-xD7-10-x2212-34-xA0- -xA0-6-08-xD7-1018HzSo- the answer is option -A-

The maximum frequency of X-rays produced in an X-ray tube at an operating voltage of 25kV is (h=6.63×10−34Js and c=3×108 m/s)4.06×1018Hz9.2×1018Hz6.08×1018Hz3.02×1018Hz

Operating voltage =hνeoperating voltage×eh=νν=25×103×1.6×10−196.63×10−34 =6.08×1018HzSo, the answer is option (A).

The maximum frequency of X-rays produced in an X-ray tube at an operating voltage of 25kV is
(h $= 6.63 \times 10^{- 34} J s$ and $c = 3 \times 10^{8}$ m/s)

$4.06 \times 10^{18} H z$
$9.2 \times 10^{18} H z$
$6.08 \times 10^{18} H z$
$3.02 \times 10^{18} H z$

$O p e r a t i n g v o l t a g e = \frac{h ν}{e}$

$\frac{o p e r a t i n g v o l t a g e \times e}{h} = ν$

$ν = \frac{25 \times 10^{3} \times 1.6 \times 10^{- 19}}{6.63 \times 10^{- 34}}$

$= 6.08 \times 10^{18} H z$
So, the answer is option (A).