Question

Using a pen comeletely bubble in yr responges as eoted below De net makn y stey marks cn this deet, Comgletely ersse sy mistakes. Energy in the quantum mechanical model of the atom, what does an atomic orbital represent? er and Version Bl in bubble): A The exact location of an atom's electrons. The circular paths the electrons follow as they circle around the nucleus. 1 B. C The probability density of an atom's electrons. D. Both options A andB E. Both options A and C 7. Calculate the energy of the violet light emitted by a hydrogen atom with a wavelength of 4.100x 10 m. A 2.75x 10 19J B. 3.64 x 10 19J C 5.46 x 1019 D. 1.83x 10 19J E. 4.85 x 10 19 J PSU (print aumber १3 8. An FM radio station broadcasts electromagnetic radiation at a frequency of 101.2 s. The wavelength of this radiation is m. A. 2.964 x 10 B. 2.964 C. 3.036 x 1016 D. 3.036 x 1010 E. 0.3373 9. How many different values of / are possible in the 4d sublevel? A. 2 В. 1 C. 3 D. 5 E. The 4d sublevel does not exist 10. Each of the following sets of quantum numbers is supposed to specify an orbital. Choose the one set of quantum numbers that DOES NOT contain an error. A. n 2,/ 2, m/-1 B. n-2,/= 2, m/= -3 C. n 3,/=2, m/=-3 D. n=4,/=3, mj= +2 E. n=4,/= 2, ml= +4 oooo evels as noted in the ton?

Answer 1

6. Atomic orbital is defined as the most probable location to find electrons. But according to Heisenberg's Uncertainty Principle, the exact location of a microscopic particle moving with high speed, like electron, can not measured with zero uncertainty.

A atomic orbital can be occupied with the maximum of 2 electrons. They revolve the nucleus in circular paths(according to Bohr atomic model of quantum mechanics).

But orbital does not represent the probability density of an atom's electrons.

Therefore Option B is correct here.

7. & 8.

A Wavelength of the emitted violet light, a= 4.10 x 107 m (given). From Planck - Einstein relation, enery (E) associated with a light, having frequency 2, is . E= hr (2= Planck's constant = 6-626 X10-34 J:s ) or, E= be Ex-C ; c= speed of light - 3x108 m/s. I 6.626x10 34 x3x108 J.S.xm.5t 4.10 x10-7 m & Hence, E= 4.85 x 101 ) Option ® o E- -19 Here, given, frequency of the electro- magnetic radiation , V = 101.25 Therefore, wavelength of the radiation, 1- » § → 1-$] a = 3x108 mist 101.2 5 So a= 2964426.88 m. a ł 2.964 . 106 m) Option ④

9. For any sublevel only one value of l (orbital quantum number) is possible. For any d-sublevel, l=2. Therefore, the correct option is B.

Note: For s-sublevel, l=0 ; For p-sublevel, l=1 etc. For a sublevel with orbital quantum number, l, there are (2l+1) number of orbitals present. Each orbital has a different m_l(magnetic quantum number) value, ranging from +l to 0 to -l.

10. Option A and B : For n=2, only s and p-sublevels are possible; for which either l=0(for s) or l=1(for p) and can not be greater than 1. So, these two are incorrect.

Option C : For n=3, l=2 is possible (3d-sublevel). But m_l can have values +2,+1,0,-1 and -2. It cannot have a value greater than +l(here +2) or lower than -l(here -2). So, this is also incorrect.

Option D : For n=4, l=3 is possible (4f-sublevel). m_l can have values +3,+2,+1,0,-1,-2 and -3. Here m_l=+2. Therefore, this option is Correct.

Option E : n=4, l=2 is for 4d-sublevel. But m_{​​​​​​l} can not have a value of +4(greater than +2). So, this also is incorrect.