In an X- ray tube- the accelerating voltage is 20 kV- Two targets A and B are used one by one- For A the wavelength of the K- line is 62 pm- For B the wavelength of the L- line is 124 pm- The energy of the B ion with vacancy in M shell is 5-5 keV higher than atom of B-Take hc -12400 eVA-

(1.) As we know that, λ _min=hcKE of accelerated electron λ _min =hceV=1240020 × 10^3A=0.62 A ∴λ _min=62 pm nbsp; (2.) Since nbsp;λ _min=62 pm, so K_α ph ...

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Byju's Answer

Standard XII

Physics

Characteristic Spectrum

In an X- ray ...

Question

In an $X -$ ray tube, the accelerating voltage is $20 kV$ . Two targets $A$ and $B$ are used one by one. For $A$ the wavelength of the $K_{α}$ line is $62 pm$ , For $B$ the wavelength of the $L_{α}$ line is $124 pm$ . The energy of the $B$ ion with vacancy in $M$ shell is $5.5 keV$ higher than atom of $B$ .
$[Take h c = 12400 eV ˚ A]$

Value of

λ_{m i n}

0.62 ˚ A

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A

will emit

K_{α} -

photon

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B

will emit

L -

photons.

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Minimum wavelengths

(in ˚ A)

of the characteristic

X -

ray that will be emitted by

B

0.8 ˚ A

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Solution

The correct option is D Minimum wavelengths $(in ˚ A)$ of the characteristic $X -$ ray that will be emitted by $B$ is $0.8 ˚ A$
$(1.)$
As we know that,
$λ_{m i n} = \frac{h c}{KE of accelerated electron}$

$λ_{m i n} = \frac{h c}{e V} = \frac{12400}{20 \times 10^{3}} ˚ A = 0.62 ˚ A$

$∴ λ_{m i n} = 62 pm$

$(2.)$
Since $λ_{m i n} = 62 pm$ , so $K_{α} -$ photon from $A$ will not be obtained.

$(3.)$
As we know that $L$ - photon will emit when electron of $L$ - shell removed.

Given that, the energy of $B$ ion with vacancy in $M$ shell is $5.5 keV$ more than the energy of $B$ atom, which means the energy required to ionize $B$ from $M$ shell is

$E_{M \to \infty} = 5.5 keV$ .

Also given that, $λ_{L_{α}} = 124 pm$ ,

$\Rightarrow E_{L \to M} = \frac{h c}{λ_{L_{α}}} = \frac{12400 eV ˙ A}{1.24 ˙ A}$

$\Rightarrow E_{L \to M} = 10 keV$ .

$∴$ The total energy required to remove electron from $L -$ shell is

$E_{L \to \infty} = E_{L \to M} + E_{M \to \infty}$

$\Rightarrow E_{L \to \infty} = 10 keV + 5.5 keV = 15.5 keV$

As the energy of incoming electron is $20 keV > 15.5 keV,$

The $L -$ shell electron can be removed.

Hence, $L -$ photon can be obtained from $B$ .

$(4.)$
The minimum wavelength will correspond to the transition from $\infty$ to $L -$ shell.

$λ = \frac{12400}{15.5 \times 10^{3}} ˚ A = 0.8 ˚ A$

Hence, options $(A), (C)$ and $(D)$ are correct.

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In an X− ray tube, the accelerating voltage is 20 kV. Two targets A and B are used one by one. For A the wavelength of the Kα line is 62 pm, For B the wavelength of the Lα line is 124 pm. The energy of the B ion with vacancy in M shell is 5.5 keV higher than atom of B. [Take hc=12400 eV˚A]