Atomic Structure - NEET/JEE /KEAM

Explanation
The wavelength of a photon can be determined using the equation E = hc/λ, where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the photon. Rearranging the equation to solve for λ, we get λ = hc/E. Plugging in the given values of E = 3.03 × 10-19 J, h = 6.626 × 10-34 J·s, and c = 3 × 108 m/s, we can calculate the wavelength to be 656 nm.

Explanation
In photo-electron emission, the energy of the emitted electron is smaller than that of the incident photon. This is because when a photon interacts with an electron, only a portion of its energy is transferred to the electron. The remaining energy is lost in the form of heat or other processes. Therefore, the energy of the emitted electron is less than the energy of the incident photon.

Explanation
Davisson and Germer showed experimentally that electrons have wave properties through their famous electron diffraction experiment in 1927. They directed a beam of electrons at a nickel crystal and observed a diffraction pattern, similar to the diffraction of light waves passing through a narrow slit. This confirmed the wave-particle duality of electrons, supporting the concept proposed by Louis de Broglie that particles like electrons can exhibit wave-like behavior. This experiment played a crucial role in the development of quantum mechanics and our understanding of the nature of matter.

Explanation
Cathode rays are not electromagnetic waves because they are actually streams of electrons. They possess kinetic energy because they are moving particles. They can produce heat and mechanical pressure when they collide with other objects, but they themselves are not electromagnetic waves.

Explanation
According to the Bohr theory, the energy of a photon emitted or absorbed during a transition in the hydrogen atom is given by the equation E = -13.6 eV * (1/n^2). The transition with the least energetic photon will have the smallest change in energy. In this case, the transition from n=6 to n=5 has a smaller change in energy compared to the other options, resulting in the least energetic photon.

Explanation
The energy of an electron in an orbit of a hydrogen atom is given by the equation E = -13.6/n^2 eV, where n is the principal quantum number. The first excited state corresponds to n = 2. Plugging this value into the equation, we get E = -13.6/2^2 = -3.4 eV. Since the question asks for the kinetic energy (KE) of the same orbit, the KE would also be -3.4 eV.

Explanation
According to Bohr's theory, the energy of an electron in an atom is quantized and can only exist in certain discrete energy levels. The energy required for an electron to be emitted from a particular energy level is given by the formula ΔE = E_initial - E_final, where E_initial is the energy of the initial state and E_final is the energy of the final state. In this case, the electron is being emitted from the n=2 state, so E_initial = -13.6 eV (ground state energy of hydrogen atom) and E_final = 0 eV (emitted electron is completely free). Therefore, ΔE = -13.6 eV - 0 eV = -13.6 eV. Since energy cannot be negative, the absolute value of ΔE is taken, giving 13.6 eV. However, the question asks for the energy required, so the answer is 13.6 eV.

Explanation
The radius for the first excited state (n=2) orbit of the hydrogen atom can be calculated using the formula for the Bohr radius, which is given by r = 0.529 * n^2/Z, where r is the radius, n is the principal quantum number, and Z is the atomic number. In this case, n=2 and Z=1 for hydrogen. Plugging in these values into the formula, we get r = 0.529 * 2^2/1 = 0.529 * 4 = 2.12 A. Therefore, the correct answer is 2.12.

Explanation
In the ground state, the radius of an atom is determined by the size of its electron cloud. As the atomic number increases, the number of protons in the nucleus also increases, leading to a stronger attraction between the protons and the electrons. This stronger attraction causes the electron cloud to be pulled closer to the nucleus, resulting in a smaller radius. Since Li2+ has an atomic number of 3, it has one less electron than hydrogen. Therefore, the radius of Li2+ in a similar state would be smaller than that of hydrogen. Among the given options, 0.17 A is the smallest radius, indicating that it is the correct answer.

Explanation
The ionization potential for an atom is the energy required to remove an electron from the atom. In the case of hydrogen, the ionization potential is 13.6 eV. When hydrogen loses an electron, it becomes a hydrogen ion (H+). However, in the given question, we are asked about the ionization potential for He+ (helium ion). He+ is a helium atom that has lost one electron. Since helium has two electrons, removing one electron requires more energy than removing one electron from hydrogen. Therefore, the ionization potential for He+ is higher than that of hydrogen. Among the given options, the only value that is higher than 13.6 eV is 54.4 eV, so that is the correct answer.

Explanation
The given answer, rn^2, is the correct formula for calculating the radius of the nth orbit of a hydrogen atom. This formula is derived from the Bohr model of the atom, which states that the radius of each orbit is proportional to the square of the principal quantum number, n. Therefore, multiplying the radius of the first orbit, r, by n^2 gives us the radius of the nth orbit.

Explanation
The spectrum of He is expected to be similar to that of Li+ because both He and Li+ have similar electron configurations. He has two electrons in its 1s orbital, while Li+ has only one electron in its 1s orbital. This similarity in electron configuration results in similar energy levels and transitions, leading to similar spectra.

Explanation
Two electrons occupying the same orbital are distinguished by their spin quantum number. The spin quantum number describes the intrinsic angular momentum of an electron and can have two possible values: +1/2 or -1/2. This means that two electrons in the same orbital must have opposite spin values, allowing them to occupy the same space without violating the Pauli exclusion principle. The other quantum numbers (principal, magnetic, and azimuthal) are used to describe the energy level, orientation, and shape of the orbital, but they do not differentiate between electrons in the same orbital.

Explanation
The set of quantum numbers for an electron in an atom describes its energy level, orbital shape, orientation, and spin. The first quantum number, n, represents the energy level or shell of the electron. The second quantum number, l, represents the orbital shape or subshell of the electron. The third quantum number, ml, represents the orientation or magnetic quantum number of the electron within the subshell. The fourth quantum number, ms, represents the spin of the electron. In the given options, the correct set of quantum numbers for the valence electron of rubidium (Z=37) is 5,0,0,+1/2. This indicates that the valence electron is in the fifth energy level, s orbital (l=0), with no specific orientation (ml=0), and a spin of +1/2.

Explanation
The fourth energy level of an atom contains 4 sublevels: s, p, d, and f. The s sublevel has 1 orbital, the p sublevel has 3 orbitals, the d sublevel has 5 orbitals, and the f sublevel has 7 orbitals. Adding up the number of orbitals in each sublevel, we get 1 + 3 + 5 + 7 = 16. Therefore, the total number of atomic orbitals in the fourth energy level of an atom is 16.