If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is $0.01 n m$ , then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately
$(h c = 12400 e V A, h = 6.63 \times 10^{- 34}$ in MKS, mass of electron $= 9.1 \times 10^{- 31} k g)$

Question

If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is

0.01 n m

, then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately

(h c = 12400 e V A, h = 6.63 \times 10^{- 34}

in MKS, mass of electron

= 9.1 \times 10^{- 31} k g)

Toppr Admin · Accepted Answer

The shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is 0-01nm-xA0-So- -x3BB-min-12400V-x21D2-V-12400-01-124000volts

If the shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is 0.01nm, then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately(hc=12400eVA,h=6.63×10−34 in MKS, mass of electron=9.1×10−31kg)0.350.035351350

The shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is 0.01nm. So, λmin=12400V⇒V=12400.01=124000volts

The shortest wavelength of the continuous X-ray spectrum coming out of a Coolidge tube is $0.01 n m$ .

So, $λ_{m i n} = \frac{12400}{V} \Rightarrow V = \frac{1240}{0.01} = 124000 v o l t s$