The quantum numbers of four electrons are given below :
I. $$n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$$
II. $$n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$$
III. $$n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$$
IV. $$n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$$
The correct decreasing order of energy of these electrons is
Given below are two statements:
Statement I: The Balmer spectral line for H atom with lowest energy is located at $$\frac{5}{36} R_H \mathrm{~cm}^{-1}$$.
($$\mathrm{R}_{\mathrm{H}}=$$ Rydberg constant)
Statement II: When the temperature of blackbody increases, the maxima of the curve (intensity and wavelength) shifts to shorter wavelength.
In the light of the above statements, choose the correct answer from the options given below:
The energy of an electron in the ground state $$\mathrm{(n=1)}$$ for $$\mathrm{He}^{+}$$ ion is $$\mathrm{-x} \mathrm{~J}$$, then that for an electron in $$\mathrm{(n=2)}$$ state for $$\mathrm{Be}^{3+}$$ ion in $$\mathrm{J}$$ is
Match List I with List II.
| List I (Quantum Number) |
List II (Information provided) |
||
|---|---|---|---|
| A. | $$\mathrm{m_l}$$ | I. | Shape of orbital |
| B. | $$\mathrm{m_s}$$ | II. | Size of orbital |
| C. | I | III. | Orientation of orbital |
| D. | n | IV. | Orientation of spin of electron |
Choose the correct answer from the options given below :