Question

In He- Ne laser, metastable state exists in:
$(A)$ He
$(B)$ Ne
$(C)$ Both $$\left(A\right)$$ and$$\left(B\right)$$ .
$(D)$ Neither He nor Ne

Answer 1

Hint: He- Ne laser is the first gas laser, as well as the first CW laser that was brought into the operation in $$1961$$ . The He-Ne laser produced light at $$1152nm$$ . By selecting the proper resonator mirrors the laser outputs can be derived. Remember that the always energy is transferred from lower to higher that is a metastable state.

Complete step by step answer:
The helium-neon laser consists of a long and narrow discharge tube that is filled with $$1mmHg$$ of He and $$0.1mmHg$$ of the Ne. The neon atoms provide the energy levels for the laser transmissions. The helium atoms are not directly involved in the transitions, they provide an efficient mechanism for the excitation of the neon atoms. At the ends of the glass, the cell is closed by the mirrors. One of the mirrors is 100 percent reflective.
When the discharge starts the high energy electrons produced in the tube collide with the gas phase atoms and excite them to the excited states.
In transition in the He-Ne laser, the metastable states are at Helium that nearly coincides with the energy levels of the neon atoms. This is because the cell contains a higher concentration of helium atoms that collide with the neon atoms. These collisions generate helium atoms in a very excited state that is a metastable state.

Therefore, the correct option is option $(A)$.

Note: There are three major lazer transitions of laser in the helium-neon laser. The helium-neon lasers are extensively used in laboratories where highly coherent and monochromatic lasers are needed. It is widely used in the devices of supermarket scanners, range finders, and printers, etc.