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When a transition of electron in He^(+) ...

When a transition of electron in `He^(+)` takes place from `n_(2)" to "n_(1)` then wave number in terms of Rydberg constant R will be
`("Given "n_(1)+n_(2)=4, n_(2)-n_(1)=2)`

A

`(3R)/(4)`

B

`(8R)/(9)`

C

`(32R)/(9)`

D

(24R)/(9)`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the wave number when an electron transitions from level \( n_2 \) to \( n_1 \) in the helium ion \( He^+ \). The given conditions are: 1. \( n_1 + n_2 = 4 \) 2. \( n_2 - n_1 = 2 \) ### Step 1: Solve for \( n_1 \) and \( n_2 \) We have two equations: 1. \( n_1 + n_2 = 4 \) (Equation 1) 2. \( n_2 - n_1 = 2 \) (Equation 2) To find \( n_1 \) and \( n_2 \), we can add these two equations together: \[ (n_1 + n_2) + (n_2 - n_1) = 4 + 2 \] This simplifies to: \[ 2n_2 = 6 \] Dividing both sides by 2 gives: \[ n_2 = 3 \] Now, substitute \( n_2 \) back into Equation 1 to find \( n_1 \): \[ n_1 + 3 = 4 \] So, \[ n_1 = 1 \] ### Step 2: Use the wave number formula The wave number \( \bar{\nu} \) for a transition in a hydrogen-like atom is given by the formula: \[ \bar{\nu} = R z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where: - \( R \) is the Rydberg constant, - \( z \) is the atomic number (for \( He^+ \), \( z = 2 \)), - \( n_1 = 1 \) and \( n_2 = 3 \). ### Step 3: Substitute the values into the formula Now, substituting \( n_1 \), \( n_2 \), and \( z \) into the wave number formula: \[ \bar{\nu} = R (2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \] Calculating \( z^2 \): \[ 2^2 = 4 \] Now substituting this into the equation: \[ \bar{\nu} = R \cdot 4 \left( 1 - \frac{1}{9} \right) \] Calculating \( 1 - \frac{1}{9} \): \[ 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \] Now substituting this back into the wave number equation: \[ \bar{\nu} = R \cdot 4 \cdot \frac{8}{9} \] Calculating this gives: \[ \bar{\nu} = \frac{32R}{9} \] ### Final Answer The wave number in terms of the Rydberg constant \( R \) is: \[ \bar{\nu} = \frac{32R}{9} \]
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Knowledge Check

  • The transition of an electron from n_(2)=5,6 ….. To n_(1) =4 gives rise to

    A
    Pfund series.
    B
    Lyman series.
    C
    Paschen series
    D
    Brackett series.

  • the only electron in the hydrogen atom resides under ordinary conditions in the first orbit. When energy is supplied the electron moves to higher energy orbit depending on the amount of energy absorbed. It emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second. Similarly, Paschen, Breakett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively (as shown in figure). Maximum number of different lines produced when electron jump from nth level to ground level is equal to (n(n-1))/(2). For example in the case of n=4, number of lines produced is 6.(4to3,4to2,4to1,3to2,3to1,2to1). When an electron returns from n_(2) "to"n_(1) state, the number of different lines in the spectrum will be equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(1), then the difference may be expressed in terms of energy of photon as: E_(2)-E_(1)=DeltaE,lambda=(hc)/(DeltaE),DeltaE=hv(v-"frequency") Since h and c are constants DeltaE corresponds to definite energy: thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula: barv=RZ^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) where R is a Rydberg constant (R=1.1xx10^(7)m^(-1)). (i) First line of a series : It is called line of longest wavelength of line of smallest energy'. (ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy. wave number of the first line of Paschen series in Be^(3+) ion is :

    A
    `(7R)/(16)`
    B
    `(7R)/(144)`
    C
    `(7R)/(9)`
    D
    `(R)/(144)`

  • The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +(1)/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) The wave number of electromagnetic radiation emitted during the transition of electron in between two levels of Li ^(2+) ion whose principal quantum numbers sum is 4 and difference is 2 is:

    A
    `3.5R`
    B
    `4R`
    C
    8 R
    D
    `8//9R`

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    The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2). For exampe, in the case of n=4, number of lines produced is 6 (4 to 3,4 t 2,4 to 1,3to 2,3 to 1,2 to 1). When an electron returns from n_(2) and n_(1) state, the number of lines in the spectrum will be equal to (n _(2)-n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E _(2) to energy level having energy E _(1), then the difference may be expressed in terms of energy of photon as : E_(2) -E_(1) =Delta E, lamda= (hc) /(Delta E) Since h and c are constants, Delta E corresponds to definite energy, thus each transition from one energy level to another will produce a light of definitewavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula vecv=R((1)/(n _(1)^(2))-(1)/(n _(2) ^(2))), where R is a Rydberg's constant (R=1.1 xx 10 ^(7)m^(-1)) The energy photon emitted corresponding to transition n =3 to n =1 is [h=6 xx 10 ^(-34)J-sec]

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