The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is x times the frequency for the corresponding line in the case of H atom, where x is

Question

The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is x times the frequency for the corresponding line in the case of H atom, where x is (A) √2 (B) 1 / 2 √2 (C) 2 √2 (D) 1 / 2

Tardigrade Admin · Accepted Answer

Rydberg's constant determines the frequencies. We have R propto m. So, modified R for positronium atom is half of H atom. Hence, frequencies are reduced to half.

Q. The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is $x$ times the frequency for the corresponding line in the case of $H$ atom, where $x$ is

$2$

$1/2 2$

$22$

$1/2$

Rydberg's constant determines the frequencies. We have $R \propto m$ .
So, modified $R$ for positronium atom is half of $H$ atom.
Hence, frequencies are reduced to half.

Q. The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is x times the frequency for the corresponding line in the case of H atom, where x is

2​

1/22​

22​

1/2

Rydberg's constant determines the frequencies. We have R∝m. So, modified R for positronium atom is half of H atom. Hence, frequencies are reduced to half.

Q. The frequency of emission line for any transition in positronium atom (consisting of a positron and an electron) is $x$ times the frequency for the corresponding line in the case of $H$ atom, where $x$ is

$2$

$1/2 2$

$22$

$1/2$

Rydberg's constant determines the frequencies. We have $R \propto m$ .
So, modified $R$ for positronium atom is half of $H$ atom.
Hence, frequencies are reduced to half.