Atoms practice · Mixed difficulty

Option A is correct because increasing the potential difference between the anode and filament increases the kinetic energy, and thus the velocity, of the electrons.

Option D ($180^\circ$) is correct because if the impact parameter is zero, the alpha particle will retrace its path, resulting in a deviation of 180 degrees.

Option A ($2.92 \times {10^{15}}\,{\text{Hz}}$) is correct because it is calculated using the Rydberg formula for hydrogen-like ions, relating frequency to the difference in inverse squares of principal quantum numbers.

Option A (cobalt) is correct because its atomic number (Z=27) satisfies the approximate Moseley's law for K-alpha X-ray wavelength.

Option A (Electrons) is correct because the intensity of X-rays is directly related to the number of electrons striking the target material.

Option C (822 nm) is correct because the series limit for the Paschen series is calculated using the Rydberg formula with n1=3 and n2=infinity.

Option B ($10^{-24}$) is correct because it is calculated using the Heisenberg uncertainty principle.

Option D (An electronic transition of the target atom) is correct because characteristic X-rays are produced when electrons transition between energy levels in the target atom.

Option A ($T_n \propto \frac{{n^3}}{{Z^2}}$) is correct because the time period is proportional to n cubed divided by Z squared, derived from Bohr's model.

Option B ($3 \times {10^8}m/s$) is correct because X-rays are electromagnetic waves and travel at the speed of light.

Option C is correct because Bohr's third postulate correctly states the quantization of angular momentum, but the linear momentum is not directly quantized in the same way.

Option C ($5 \mathrm{~Hz}$) is correct because it is calculated using the formula for the frequency of a standing wave in a string fixed at both ends.

Option D (Hyperbolic) is correct because the path of scattered alpha particles in Rutherford's experiment is hyperbolic due to the repulsive Coulomb force between the alpha particles and the nucleus.

Option A ($2\;\mathop A\limits^ \circ$) is correct because X-rays with shorter wavelengths have greater penetrating power.

Option B is correct because both the assertion and reason are true, but the reason does not correctly explain the assertion. The deviation is small because the alpha particle is far from the nucleus, and the impact parameter is defined as the perpendicular distance.

Option C ($U = 2E$) is correct because the potential energy is twice the total energy in magnitude and has the same sign.

Option B (Mechanical energy) is correct because under a conservative force, the total mechanical energy (kinetic + potential) is conserved.