An X-Ray tube operates a p-d- of 12400 V- The wavelength of the highest frequency X-Rays emitted by this tube is nearly-hc-1240 eV-nm-12- 4 A-o-1-24 A-o-2-48 A-o-1 A-o-

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Toppr Admin · Accepted Answer

Energy-p-d-xD7-e-12400V-xD7-e-12400-xA0-eVEnergy-hc-x3BB-x3BB-hcenergy -xA0-124012400-xA0-nm-0-1-xA0-nm-1-xA0-A0

An X-Ray tube operates at a p.d. of $12400 V$ . The wavelength of the highest frequency X-Rays emitted by this tube is nearly:( $h c = 1240 e V - n m$ )
$12. 4 A^{o}$
$1.24 A^{o}$
$2.48 A^{o}$
$1 A^{o}$

$E n e r g y = p . d \times e$
= $12400 V \times e$
$= 12400 e V$
$E n e r g y = \frac{h c}{λ}$
$λ = \frac{h c}{e n e r g y}$
$= \frac{1240}{12400} n m$
$= 0.1 n m$
$= 1 A^{0}$

An X-Ray tube operates at a p.d. of 12400 V. The wavelength of the highest frequency X-Rays emitted by this tube is nearly:(hc=1240 eV−nm)12.4Ao1.24Ao2.48Ao1Ao

Energy=p.d×e=12400V×e=12400 eVEnergy=hcλλ=hcenergy =124012400 nm=0.1 nm=1 A0

An X-Ray tube operates at a p.d. of $12400 V$ . The wavelength of the highest frequency X-Rays emitted by this tube is nearly:( $h c = 1240 e V - n m$ )
$12. 4 A^{o}$
$1.24 A^{o}$
$2.48 A^{o}$
$1 A^{o}$

$E n e r g y = p . d \times e$
= $12400 V \times e$
$= 12400 e V$
$E n e r g y = \frac{h c}{λ}$
$λ = \frac{h c}{e n e r g y}$
$= \frac{1240}{12400} n m$
$= 0.1 n m$
$= 1 A^{0}$