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Hydrogen is the simplest atom of nature....

Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr's, this electron moves in a stationary orbit, if emits no electromagnetic radiation. The angular momentum of the electron is quantized , i.e., `mvr = (nh//2 pi)`, where `m =` mass of the electron , `v =`velocity of the electron in the orbit , `r =`radius of the orbit , and `n = 1, 2, 3`.... When transition takes place from `Kth` orbit to `Jth` orbit, energy photon is emitted. If the wavelength of the emitted photon is `lambda`.

we find that `(1)/(lambda) = R [(1)/(J^(2)) - (1)/(K^(2))]`, where `R` is
Rydberg's constant.
On a different planed, the hydrogen atom's structure was somewhat different from ours. The angular momentum of electron was `P = 2n (h//2 pi)`. i.e., an even multipal of `(h//2 pi)`.
Answer the following questions regarding the other planet based on above passage:
In our world, the velocity of electron is `v_(0)` when the hydrogen atom is in the ground state on the other planet should be

Text Solution

Verified by Experts

The correct Answer is:
B

On our planed `v_(0) = (e^(2))/(2 epsilon_(0) n h)`
On other planet `v = (e^(2))/(2 epsilon_(0) (2 n)h) = (v_(0))/(2)`
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Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr's, this electron moves in a stationary orbit, if emits no electromagnetic radiation. The angular momentum of the electron is quantized , i.e., mvr = (nh//2 pi) , where m = mass of the electron , v = velocity of the electron in the orbit , r = radius of the orbit , and n = 1, 2, 3 .... When transition takes place from Kth orbit to Jth orbit, energy photon is emitted. If the wavelength of the emitted photon is lambda . we find that (1)/(lambda) = R [(1)/(J^(2)) - (1)/(K^(2))] , where R is Rydberg's constant. On a different planed, the hydrogen atom's structure was somewhat different from ours. The angular momentum of electron was P = 2n (h//2 pi) . i.e., an even multipal of (h//2 pi) . Answer the following questions regarding the other planet based on above passage: The minimum permissible radius of the orbit will be

Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr's, this electron moves in a stationary orbit, if emits no electromagnetic radiation. The angular momentum of the electron is quantized , i.e., mvr = (nh//2 pi) , where m = mass of the electron , v = velocity of the electron in the orbit , r = radius of the orbit , and n = 1, 2, 3 .... When transition takes place from Kth orbit to Jth orbit, energy photon is emitted. If the wavelength of the emitted photon is lambda . we find that (1)/(lambda) = R [(1)/(J^(2)) - (1)/(K^(2))] , where R is Rydberg's constant. On a different planet, the hydrogen atom's structure was somewhat different from ours. The angular momentum of electron was P = 2n (h//2 pi) . i.e., an even multiple of (h//2 pi) . Answer the following questions regarding the other planet based on above passage: In our world, the ionization potential energy of a hydrogen atom is 13.6 eV . On the other planet, this ionization potential energy will be

Knowledge Check

  • The orbital angular momentum of an electron in 2s -orbital is

    A
    Zero
    B
    One
    C
    Two
    D
    Three
  • According to Bohr's theory, the electronic energy of H-atom in Bohr's orbit is given by

    A
    `E_(n)=-(2.18xx10^(-19)xxZ)/(2n^(2))J`
    B
    `E_(n)=-(2.179xx10^(-18)xxZ^(2))/(n^(2))J`
    C
    `E_(n)=-(21.79xx10^(-18)xxZ)/(2n^(2))J`
    D
    `E_(n)=-(21.8xx10^(-21)xxZ^(2))/(n^(2))J`
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