If the kinetic energy of a particle is increased to 16 times, the percentage decrease in de Broglie wavelength of particle is
Using \(\lambda = \frac{h}{\sqrt{2mK}}\), when kinetic energy \(K\) becomes \(16K\), the new wavelength becomes \(\lambda' = \frac{\lambda}{4}\). The percentage decrease is \(\frac{\lambda - \lambda/4}{\lambda} \times 100\% = 75\%\).
The stopping potential in the photoelectric experiment is \( 1.6\text{ V} \). The maximum kinetic energy of photoelectrons emitted is
The maximum kinetic energy of photoelectrons is given by \( K_{\text{max}} = e V_0 \). Substituting the values: \( K_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 1.6\text{ V} = 2.56 \times 10^{-19}\text{ J} \).
Choose the incorrect statement among the following.
Nuclides with the same atomic number \( Z \) but different neutron numbers \( N \) are called isotopes, not isotones. Thus, option C is incorrect.
Assertion (A): A particle at rest breaks into two particles of different masses. They fly off in different directions. Their de Broglie wavelengths will be different.
Reason (R): Their speed will be different.
When a particle at rest decays, conservation of momentum dictates that the two resulting particles must have equal and opposite momenta \( |p_1| = |p_2| \). Since the de Broglie wavelength is \( \lambda = h/p \), both particles must have the same wavelength, making the assertion that they are different false.
Assertion (A): Two photons having equal wavelengths have equal linear momentum.
Reason (R): When monochromatic light shows its photon character, each photon has a linear momentum \(p = \frac{h}{\lambda}\).
The linear momentum of a photon is given by \(p = h/\lambda\). If two photons have equal wavelengths \(\lambda\), then their linear momenta \(p\) must also be equal. The Reason (R) correctly states the formula and explains Assertion (A).
Assertion (A): If the accelerating potential of a X-Ray tube is increased then the characteristic wavelength decreases.
Reason (R): The cut-off wavelength for a X-Ray tube is given by \(\lambda_{\text{min}} = \frac{hc}{eV}\), where \(V\) is accelerating potential.
Characteristic wavelength depends on target material, not accelerating potential. However, if 'characteristic wavelength' in (A) refers to 'cut-off wavelength', then \(\lambda_{\text{min}} = hc/(eV)\) implies increasing \(V\) decreases \(\lambda_{\text{min}}\). Under this interpretation, (A) is true, and (R) is true and explains (A).
Assertion (A): A photon and an electron both have energy \(50\text{ eV}\). Both have different wavelengths.
Reason (R): Wavelength depends on energy and not on mass.
For a photon, wavelength is \(\lambda_p = hc/E\). For an electron, de Broglie wavelength is \(\lambda_e = h/\sqrt{2mE}\). Since their formulas are different and \(\lambda_e\) depends on mass \(m\), their wavelengths will be different for the same energy. So (A) is true. Reason (R) is false because an electron's de Broglie wavelength depends on its mass.
Assertion (A): In photoelectric effect, cathode or emitter plate is usually coated with barium oxide, barium sulphide or strontium oxide.
Reason (R): Coating prevents cathode from erosion.
Cathodes are coated with materials like barium oxide to lower their work function, enhancing photoemission efficiency. So (A) is true. Coatings can indeed prevent erosion, so (R) is also true. However, preventing erosion is not the primary reason for choosing these specific low work function materials, so (R) is not the correct explanation for (A).
Assertion (A): In photon-particle collision the total energy and total momentum are conserved.
Reason (R): The number of photons are conserved in a collision.
Concept: Conservation laws in collisions.
Formula: Total energy \(E_{\text{total}}\) and total momentum \(p_{\text{total}}\) are conserved in all collisions.
Solution: In photon-particle collisions, total energy and momentum are conserved. However, photons can be absorbed or emitted, so their number is not necessarily conserved. Hence, Assertion (A) is true, but Reason (R) is false.
Assertion (A): Cut-off wavelength of x-ray is independent of type of target metal
Reason (R): Wavelength of \(K_{alpha}\) x-ray depends upon type of target metal.
Concept: X-ray production mechanisms.
Formula: Cut-off wavelength \(\lambda_{\text{min}} = \frac{hc}{eV}\) (depends on voltage). Characteristic X-ray energy \(E = h\nu\) (depends on atomic transitions).
Solution: The cut-off (minimum) wavelength of continuous X-rays depends only on the accelerating voltage, not the target material. So (A) is true. \(K_{\alpha}\) X-rays are characteristic X-rays, whose wavelengths are specific to the target material. So (R) is true. However, (R) explains characteristic X-rays, not the cut-off wavelength; thus, (R) is not a correct explanation for (A).