In an X-ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode) at a distance d from the cathode. The target is kept at a potential V higher than the cathode resulting in emission of continuous and characteristic X-rays. If the filament current I is decreased to 1 \over 2 , the potential difference V is increased to 2V, and the separation distance d is reduced to d \over 2 , then

the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same, the cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease

the cut-off wavelength will reduce to half, and the wavelengths of the characteristic X-rays will remain the same

the cut-off wavelength as well as the wavelengths of the characteristic X-rays will remain the same

the cut-off wavelength will reduce to half, and the intensities of all the X-rays will decrease

the cut-off wavelength will become two times larger, and the intensity of all the X-rays will decrease

Atoms and Nuclei PYQ — JEE Advanced 2020 · Shift 2 · Q46 with Solution

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Correct answer: A, C

Cut-off wavelength of continuous X-rays

The minimum wavelength (cut-off wavelength) of X-rays is given by Duane–Hunt law:

\lambda_{\min} = \frac{hc}{eV}

Initially,

\lambda_{\min} = \frac{hc}{eV}

When the accelerating potential is increased from $V$ to $2V$ ,

\lambda'_{\min} = \frac{hc}{e(2V)} = \frac{1}{2}\,\frac{hc}{eV} = \frac{\lambda_{\min}}{2}

So, the cut-off wavelength reduces to half.

Thus:

Option A: first part correct
Option B: first part wrong
Option C: first part correct
Option D: first part wrong

Characteristic X-ray wavelengths

Characteristic X-rays depend on the energy level differences of the target atom, so their wavelengths depend on the material of the target and not on filament current or tube geometry.

Changing:

filament current $I \to I/2$
potential $V \to 2V$
distance $d \to d/2$

does not change the target material, hence the wavelengths of characteristic X-rays remain unchanged.

So, Option A's second part is correct.

Effect on intensity of X-rays

The filament current determines the number of electrons emitted per second. If filament current is reduced to $I/2$ , then the number of electrons striking the target per second decreases.

Hence, the intensity of emitted X-rays decreases.

Also, reducing distance $d$ does not affect the energy gained by electrons, because the kinetic energy acquired is due to the potential difference $V$ , not the separation distance. So $d$ is irrelevant here.

Since fewer electrons hit the target per second, the intensity of both continuous and characteristic X-rays decreases.

Thus, Option C is correct.

Check each option

A: Cut-off wavelength becomes half; characteristic wavelengths unchanged. ✅ Correct
B: Both cut-off and characteristic wavelengths unchanged. ❌ Wrong, because cut-off wavelength changes
C: Cut-off wavelength becomes half; intensities of all X-rays decrease. ✅ Correct
D: Cut-off wavelength becomes twice; intensity decreases. ❌ Wrong, because cut-off wavelength actually halves

Final answer

The correct options are:

\boxed{A,\ C}

This matches the stored correct answer.

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View written solutionFree

Correct answer: A, C

Cut-off wavelength of continuous X-rays

The minimum wavelength (cut-off wavelength) of X-rays is given by Duane–Hunt law:

\lambda_{\min} = \frac{hc}{eV}

Initially,

\lambda_{\min} = \frac{hc}{eV}

When the accelerating potential is increased from $V$ to $2V$ ,

\lambda'_{\min} = \frac{hc}{e(2V)} = \frac{1}{2}\,\frac{hc}{eV} = \frac{\lambda_{\min}}{2}

So, the cut-off wavelength reduces to half.

Thus:

Option A: first part correct
Option B: first part wrong
Option C: first part correct
Option D: first part wrong

Characteristic X-ray wavelengths

Characteristic X-rays depend on the energy level differences of the target atom, so their wavelengths depend on the material of the target and not on filament current or tube geometry.

Changing:

filament current $I \to I/2$
potential $V \to 2V$
distance $d \to d/2$

does not change the target material, hence the wavelengths of characteristic X-rays remain unchanged.

So, Option A's second part is correct.

Effect on intensity of X-rays

The filament current determines the number of electrons emitted per second. If filament current is reduced to $I/2$ , then the number of electrons striking the target per second decreases.

Hence, the intensity of emitted X-rays decreases.

Since fewer electrons hit the target per second, the intensity of both continuous and characteristic X-rays decreases.

Thus, Option C is correct.

Check each option

A: Cut-off wavelength becomes half; characteristic wavelengths unchanged. ✅ Correct
B: Both cut-off and characteristic wavelengths unchanged. ❌ Wrong, because cut-off wavelength changes
C: Cut-off wavelength becomes half; intensities of all X-rays decrease. ✅ Correct
D: Cut-off wavelength becomes twice; intensity decreases. ❌ Wrong, because cut-off wavelength actually halves

Final answer

The correct options are:

\boxed{A,\ C}

This matches the stored correct answer.

Atoms and Nuclei question

Atoms and Nuclei question

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Atoms and Nuclei question

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