For the ground state, the electron in the H-atom has an angular momentum \(\dfrac h{2\pi}\), according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

1. because the Bohr model gives incorrect values of angular momentum.
2. because only one of these would have a minimum energy.
3. angular momentum must be in the direction of the spin of the electron.
4. because electrons go around only in horizontal orbits.

Hint: The Bohr model does not give the direction of angular momentum.
 
Explanation: Bohr found that the magnitude of the electron's angular momentum is quantized i.e.
\(L=m v_n r_n=n\left(\frac{h}{2 \pi}\right)\)
where \({n}=1,2,3, \ldots \ldots .\) each value of \(n\) corresponds to a permitted value of the orbit radius.
\(r_n=\) Radius of \({n}^{\text {th }}\) orbit, \({v}_{\mathrm{n}}=\) corresponding speed
Bohr's model gives only the magnitude of angular momentum. The angular momentum is a vector quantity. Hence we cannot express angular momentum completely by the Bohr model. Hence it does not give correct values of the angular momentum of a revolving electron.
Hence, option (1) is the correct answer.