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Let E=(-1me^(4))/(8epsilon(0)^(2)n^(2)h^...

Let `E=(-1me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` be the energy of the `n^(th)` level of H-atom state and radiation of frequency `(E_(2)-E_(1))//h` falls on it ,

A

it will not be absorbed at all

B

some of atoms will move to the first excited state

C

all atoms will be excited to the n=2 state

D

no atoms will make a transition to the n=3 state

Text Solution

Verified by Experts

The correct Answer is:
B, D

Here, `E_(n)=(me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))`
Is the energy of nth level of hydrogen atom. If all the H-atom are in ground state, (n=1), then the radiation of frequency `(E_(2)-E_(1))//h` falling on it may not be absorbed by some of the atoms and move them to the first excited state (n=2). All atoms may not be excited to n=2 state. Further, as `(E_(2)-E_(1))//h` is sufficient only to take the atom form n=1 state to n=2 state, no atoms shall make a transition to n=3 state. Choices (b) and (d) are correct.
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Hydrogen Atom and Hydrogen Molecule The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is 1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5 R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1) Here, R_(H) is the Rydberg Constant, m_(e) is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. Determine the loest energy and the radius of the Bohr orbit of the muonic hydrogen atom. Ignore the motion of the nucleus in your calculation. The radius of the Bohr orbit of a hydrogen atom ("called the Bohr radius", a_(0)=(epsilon_(0)h^(2))/(m_(e)e^(2)prod) "is" 0.53 Å) The classical picture of an ''orbit'' in Bohr's theory has now been replaced by the quantum mechanical nation of an 'orbital'. The orbital psi 1 sigma 1s (r) for the ground state of a hydrogen atom is given by psi 1 s (r)=1/sqrt(proda_(0)^(3)) e^(r/a_(0)) where r is the distance of the electron from the nucleus and a_(0) is the Bohr radius.

Knowledge Check

  • Let E_n=-me^4/(8 epsilon^2n^2h^2) be the energy of nth level of H atom. if all the H atoms are in ground state and radiation of frequency E_2-E_1/h falls on it then It will not be absorbed at all SOme of atoms moves to first excited state. III all atoms will be excited to n=2 state. IV no atom makes a transition to the n=3 state. possible option is/are

    A
    Only I
    B
    II and III
    C
    II and IV
    D
    III and IV
  • What is the maximum degeneracy of a level of H-atom, where e^(-) has energy, E_(n) =- (Rhc)/(9) ?

    A
    1
    B
    3
    C
    5
    D
    9
  • The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is B=-(me^(4))/(8n^(2)epsilon_(0)^(2)h^(2)) (m=electron mass). If one decided to works in a frame of reference where the electron is rest, the proton would be moving around it. By similar arguments, the binding energy would be B=-(Me^(4))/(8n^(2)epsilon_(0)^(2)h^(2)) (M=proton mass) This last expression is not correct because

    A
    n would not be integral
    B
    Bohr-quantisation applies only to electron
    C
    the frame in which the electrons is at rest is not inertial
    D
    the motion of the proton would not be in circular orbits, even approximately.
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