If the Plank’s constant would be double the present value, in the Bohr’s model for Hydrogen atom,

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The energy required to ionize the Hydrogen atom in it’s ground state would be doubled

The angular momentum of the electron in the ground state would be doubled

The speed of the electron in the ground state would be halved

The radius of electronic orbit in the ground state would be four times

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} r_{0} \propto h^{2} \Rightarrow (\frac{r_{1}}{r_{2}}) = {(\frac{h_{1}}{h_{2}})}^{2} \Rightarrow (\frac{r_{1}}{4 r_{1}}) = {(\frac{h_{1}}{h_{2}})}^{2} \Rightarrow h_{2} = 2 h_{1} \\ V_{0} \propto \frac{1}{h} \Rightarrow (\frac{v_{1}}{v_{2}}) = (\frac{h_{2}}{h_{1}}) \Rightarrow (\frac{v_{1}}{\frac{v_{1}}{2}}) = (\frac{h_{2}}{h_{1}}) \Rightarrow h_{2} = 2 h_{1} \\ E \propto \frac{1}{h^{2}} \Rightarrow (\frac{E_{1}}{E_{2}}) = {(\frac{h_{2}}{h_{1}})}^{2} \Rightarrow (\frac{E_{1}}{2 E_{1}}) = {(\frac{h_{2}}{h_{1}})}^{2} \Rightarrow h_{2} = \frac{h_{1}}{\sqrt{2}} \\ L \propto h \Rightarrow (\frac{L_{1}}{L_{2}}) = (\frac{h_{1}}{h_{2}}) \Rightarrow (\frac{L_{1}}{2 L_{1}}) = (\frac{h_{1}}{h_{2}}) \Rightarrow h_{2} = 2 h_{1} \end{array}$