Question

What will be the angular momentum of an electron, if the energy of this electron in H-atom is 1.5eV (in J-s)?
A.\[1.5\times {{10}^{-34}}\]
B.\[2.1\times {{10}^{-34}}\]
C.\[3.15\times {{10}^{-34}}\]
D.\[-2.1\times {{10}^{-34}}\]

Answer 1

Hint:The angular momentum is defined as the measure of momentum of a body in rotational motion. The angular momentum is reserved for some objects that are spinning spheres, etc. the change in angular momentum is equal to the equivalent torque. If the mass of the substance decreases then the velocity or the radius must be increased to compensate for the changed values.

Complete step by step answer:
Angular momentum of an electron by means of Bohr is given by using mvr or nh/2π (wherein v is the speed, n is the orbit in which electron is, m is mass of the electron, and r is the radius of the nth orbit). Bohr’s atomic version laid down diverse postulates for the arrangement of electrons in one of a kind orbits across the nucleus. According to Bohr’s atomic version, the angular momentum of the electron orbiting around the nucleus is quantized. He in addition brought that electrons circulate simplest in the one’s orbits in which angular momentum of an electron is a critical a couple of h/2. This postulate concerning the quantisation of angular momentum of an electron became later defined by using Louis de Broglie. According to him, a moving electron in its round orbit behaves like a particle wave. The Planck regular, or Planck's constant, is the quantum of electromagnetic motion that relates a photon's power to its frequency. The Planck constant extended by a photon's frequency is the same to a photon's strength. The Planck regular is a fundamental physical quantity denoted as h.
Energy if an electron in H -atom(E)= \[\dfrac{-13.6}{{{n}^{2}}}eV\]
\[{{n}^{2}}=\dfrac{-13.6}{E}\]
\[{{n}^{2}}=\dfrac{-13.6}{-1.5}=\sqrt{9.067}\approx 3\]
As we know that, angular momentum is given by
\[p=\dfrac{nh}{2\Pi }\]
whereas, h is Planck's constant = \[6.6\times {{10}^{-34}}\] j/sec
\[p=\dfrac{6.6\times {{10}^{-34}}\times 3}{2\times 3.143}=3.15\times {{10}^{-34}}J/\sec \]
Hence the angular momentum of electron will be \[3.15\times {{10}^{-34}}\]J/sec

Hence option (3) is the correct one.

Additional Information:
The behaviour of particle waves can be considered analogously to the waves journeying on a string. Particle waves can cause status waves held below resonant conditions. On the alternative hand, we know that simplest the ones wavelengths live on which form a standing wave inside the string, that is, which have nodes on the ends. Thus, in a string, standing waves are shaped simplest whilst the total distance travelled through a wave is a crucial number of wavelengths. Hence, for any electron transferring in kth circular orbit of radius rk, the entire distance is the same as the circumference of the orbit, $2( \pi )_{rk}$

Note:
The angular momentum is constant when the objects collide without the net external force. The angular momentum is constant in this case because the two bodies of equal masses collide to each other with opposite angular impulses thereby maintaining the angular momentum. The angular momentum was discovered by Jean Buridan.