Test: Atomic Structure - IIT JAM MCQ

de Broglie has proposed that like light every elementary particle has dual nature - wave nature and particle nature. Phenomenon such as diffraction and interference are the examples of wave nature of the electromagnetic radiation.
Hence, the correct answers are : (A, B).

We know that I.E. = 13.6 [1/n₁² - l/n ₂²] ....... (1)

For the third member of Balmer series : n₁ = 2 and n₂ = 5. Putting these values into the equation (1), we have : E = 13.6 [1/4 - 1/25] = 2.856 eV.

Hence, the correct answer is : B.

 The no. of radial node(s) is given by : (n -1 -1) and the no. of angular node(s) is given by : 1 where, n and 1 are the principal and orbital(azimuthal) quantum nos.
For 3p_z orbital, n = 3 and 1=1.
Hence, the no. of radial node(s) = (3 - 1 - 1) = 1 and the angular node(s) = 1.
Hence, the correct answers are : (A, C).

Let f(x) = Ae-[ax²] ......... (1) Differentiating equation (1) with respect to x, we have:

df(x)/dx = Ae^-[ax²](-2ax) .........(2) Differentiating equation (2), we have : d²f(x)/dx² = -2aA{e^-[ax²] - 2ax} .......... (3)

The maximum will be if df(x)/dx = 0. Hence, Ae^- [ax²](-2ax) = 0 or, x = 0

If it is the maximum value, then d²f(x)/dx² will be negative at x = 0.
Now putting x = 0 at equation (3), we have : d²f(x)/dx² = -2aA.,

Hence, the correct answer is : A.

The given electronic configuration is : 3d⁷. The electron occupancy of 3d orbital is as follows :

Hence, spin, S = 5(+l/2) + 2(-l/2) = 5/2 - 2/2 = 3/2.

So, the spin multiplicity, 2S + 1 = 2.(3/2) +1= 4.

Now, L = 2(+2) + 2(+l) + 1(0) + 1(-1) + 1 (-2) = 3 = ⇒ F.

Lastly, J = |L + S | to | L - S | = | 3 + 3/2 | to | 3 - 3/2 | = 9/2, 7/2, 5/2, 3/2.

According to Hund for the more than half-filled electronic configuration ground-state will be determined by highest J value.

Hence, the ground state term symbol of the element having configuration 1s² 2s² 2p⁵ is : ^(2S+1)L_j = ⁴F_9/2. Hence, the correct answer is : B.

The energy expression for H-like system having nuclear charge Z is as follows : E_n = - 2π²mk²Z²e⁴/n²h²........ (1)

where, m = mass of electron ; Z = nuclear charge of nucleous ; n = Principal quantum number ; h = Planck's constant.
Hence, from the equation (1), we have En ∝ Z². Hence, the energy of an electron in a hydrogenic atom with nuclear charge Z varies as : Z².
Hence, the correct answer is : B.

In case of Einstein s photoelectric effect: 1. The no. of electrons ejected is dependent on the intensity of the incident radiation and it is independent of the wavelength (and hence, the frequency of the incident light).

2. The kinetic energy of the emitted electrons is linearly dependent of the wavelength (and hence, the frequency of the incident light) but it is independent of the intensity of the incident radiation.

3. Electrons ejection cannot occur at any wavelength of incident radiation, it can only occur at a particular wavelength. This particular wavelength is called Threshold wavelength. This is different for different metals.
Hence, by considering the above points, one can easily conclude that the no. of electrons ejected at a given incident wavelength depends on the intensity of the radiation.
Hence, the correct answer is : C.

If ψ is the eigen function to the Hamiltonian operator with α as the eigen value, then α must be real.
Hence, the correct answer is : D.

According to Hund degenerate orbitals in a subshell will be singly occupied first and if there is excess electrons present, the orbital will be doubly occupied. During the time of singly occupancy by electrons, electrons should have one sided spin i.e. either it will be upward or, it will be downward. On following these rule, one can easily say that C is the acceptable valence shell electronic arrangement.

Hence, the correct answer is : C.

The no. of radial i.e. orbital nodes = (n-1-1) and the no. of angular nodes = 1. where, n and 1 are the principal and azimuthal (orbital) quantum no.
According to the problem the no. of angular nodes = 2. This correspond to l = 2 i.e. d-orbital and the no. of radial nodes = 1.
Hence, n-1-1 = 1 or, n = 1-1-2 = 4.
Hence, the electron is in the 4d orbital.
Hence, the correct answer is : D.

A cubical box has three dimensions x, y and z. Hence, the total no. of co-ordinates required to specify the its state is 3. Hence, the no. of quantum numbers needed to specify the state of a cubical box are 3.
Hence, the correct answer is : C.

If S = 3/2. Then the total orientations will be = +3/2, +1/2, -1/2 and -3/2. Hence, the no. of electrons that can be accommodated in a level will be : 4.
This can also be calculated from the total spin = 2S + 1 = (2 X 3/2 + 1) = 3 + 1 = 4.
Hence, the correct answer is : C.

The given two orbitals are dxz and pz respectively. Hence, the + ve lobe of d_xz orbital will overlap with the + ve lobe of the p_z orbital and the - ve lobe of d_xz orbital will overlap with the - ve lobe of the p_z orbital. Hence, here bonding will occur through the positive overlap.

Hence, the correct answer is : A.

Given that ψ(φ) = Ae^(imφ)

Hence, the complex conjugate of ψ(φ) i.e. ψ*(φ) = Ae^(imφ)

Hence, the correct answer is : D.

The total no. of electrons in CO = 14.
The total no. of electrons in N₂ = 14.
The total no. of electrons in CN⁺ = 14.
The total no. of electrons in NO⁺ = 14.
The total no. o f electrons in O₂⁺ = 15.
CO has 14 electrons. N₂, CN^-, NO⁺ all these species have 14 electrons. But, O ₂⁺ has 15 electrons. Hence, it is not iso-electronic with CO.
Hence, the correct answer is : D.

According to Einstein the total energy of the incident photon, hv = hv₀ + K.E. or, K.E. = hv - hv₀, where hv0 represent the work function.
Hence, a plot of K.E. vs. v will be as follows:

Hence, the correct answer is : A.

 Let the given wave function ψ = (2π)^-1/2 exp(2iφ). Now, the complex conjugate of the wave function ψ* = (2π)^-1/2 exp(-2iφ).
The probability of finding the particle in a small interval dφ when the value of φ = π/2 is = ψψ*dφ = (l/2π) exp(2iφ-2iφ) dφ = dφ/2π. Hence, the correct answer is : B.

According to Heisenbergs Uncertainty Principle : 

Hence, the correct answer is : A.

The ground state energy of H-like atoms is given by :

E = - (4π²μk²Z²e⁴)/(n²h²) (In C.G.S.) (All the terms have their own meaning)

Hence, the energy term involves μZ² term.
Hence, the correct answer is : C.