JEE Advance - Chemistry (2019 - Paper 2 Offline - No. 3)
The ground state energy of hydrogen atom is $$-$$13.6 eV. Consider an electronic state $$\psi $$ of He+ whose energy, azimuthal quantum number and magnetic quantum number are $$-$$3.4 eV, 2 and 0, respectively.
Which of the following statement(s) is(are) true for the state $$\psi $$?
Which of the following statement(s) is(are) true for the state $$\psi $$?
It is a 4d state
The nuclear charge experienced by the electron in this state is less than 2e, where e is the magnitude of the electronic charge
It has 2 angular nodes
It has 3 radial nodes
Explanation
Given, ground state energy of hydrogen atom = $$-$$13.6 eV
Energy of He+ = $$-$$3.4 eV, Z = 2
Energy of He+, E = $$ - {{13.6 \times {Z^2}} \over {{n^2}}}eV$$
$$ - 3.4eV = {{ - 13.6 \times {{(2)}^2}} \over {{n^2}}}$$
$$ \Rightarrow $$ n = $$\sqrt {{{ - 13.6 \times 4} \over {34}}} $$ $$ \Rightarrow $$ n = 4
Given, azimuthal quantum number (l) = 2 (d $$-$$ subshell)
Magnetic quantum number (m) = 0
$$ \therefore $$ Angular nodes (l) = 2
Radial node = n $$-$$ l $$-$$ 1 = 4 $$-$$ 2 $$-$$ 1 = 1
nl = 4d state
Hence, options (a), (c) are correct.
Energy of He+ = $$-$$3.4 eV, Z = 2
Energy of He+, E = $$ - {{13.6 \times {Z^2}} \over {{n^2}}}eV$$
$$ - 3.4eV = {{ - 13.6 \times {{(2)}^2}} \over {{n^2}}}$$
$$ \Rightarrow $$ n = $$\sqrt {{{ - 13.6 \times 4} \over {34}}} $$ $$ \Rightarrow $$ n = 4
Given, azimuthal quantum number (l) = 2 (d $$-$$ subshell)
Magnetic quantum number (m) = 0
$$ \therefore $$ Angular nodes (l) = 2
Radial node = n $$-$$ l $$-$$ 1 = 4 $$-$$ 2 $$-$$ 1 = 1
nl = 4d state
Hence, options (a), (c) are correct.
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