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In hydrogen spectrum wave number of diff...

In hydrogen spectrum wave number of different lines is given by `1/lambda=R_(H)[1/n_(i)^(2)-1/n_(f)^(2)]` where `R_(H)=1.090678xx10^(7)m^(-1)` The wavelength of first line of Lyman series would be

A

`122xx10^(-7) m`

B

`9.1176xx10^(-8)m`

C

`12.2xx10^(-7)m`

D

`1.22xx10^(-7)m`

Text Solution

Verified by Experts

The correct Answer is:
D
For first line in Lyman series
`n_(i)=1,n_(f)=2`
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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. Last line of breakett series for H-atom has wavelength lambda_(1)"Å" and 2nd line of Lyman series has wavelength lambda_(2)"Å" then:

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 :

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. Consider the following statements (i) Spectral lines of He^(+) ion belonging to Balmer series are not in visible range. (ii) In the Balmer series of H-atom maximum lines are in ultraviolet ragion. (iii) 2nd line of Lyman series of He^(+) ion has energy 48.35 eV The above statement (i), (ii), (iii) respectively are (T=True, F=False):

When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2)) Where, m_(e)= rest mass of electron DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1) Where R_(H) = 1.1 xx 10^(7)m^(-1) (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of n_(1) = 1,2,3,4,5 respectively and n_(2) =oo for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be .^(n)C_(2) . Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass (mu) . (1)/(mu) = (1)/(m_(N))+ (1)/(m_(e)) Here, m_(N)= mass of nucleus m_(e)= mass of electron Answer the following questions If the wavelength of series limit of Lyman's series of He^(+) ions is ''a'' overset(0)(A) , then what will be the wavelength of series limit of Lyman's series for Li^(2+) ion?

When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2)) Where, m_(e)= rest mass of electron DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1) Where R_(H) = 1.1 xx 10^(7)m^(-1) (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of n_(1) = 1,2,3,4,5 respectively and n_(2) =oo for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be .^(n)C_(2) . Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass (mu) . (1)/(mu) = (1)/(m_(N))+ (1)/(m_(e)) Here, m_(N)= mass of nucleus m_(e)= mass of electron Answer the following questions The emission spectrum of He^(+) involves transition of electron from n_(2) rarr n_(1) such that n_(2)+n_(1) = 8 and n_(2) -n_(1) = 4 . what whill be the total number of lines in the spectrum?

When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2)) Where, m_(e)= rest mass of electron DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1) Where R_(H) = 1.1 xx 10^(7)m^(-1) (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of n_(1) = 1,2,3,4,5 respectively and n_(2) =oo for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be .^(n)C_(2) . Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass (mu) . (1)/(mu) = (1)/(m_(N))+ (1)/(m_(e)) Here, m_(N)= mass of nucleus m_(e)= mass of electron Answer the following questions The ratio of the wavelength of the first line to that of second line of Paschen series of H-atom is

For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by bar(V)=R_(H)(1)/(n_(1)^(2))-(1)/(n_(2)^(2)) where R_(H) is a constant and n_(1) and n_(2) are integers which of the following statement (S) is / are correct 1 as wavelength decreases the lines in the series converge 2 The integer n_(1) is equal to 2 3 The ionization energy of hydrogen can be calculated from the wave number of these lines 4 The line of longest wavelength corresponds to n_(2)=3

For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by bar(V) = R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] where R_(H) is a constant and n_(1) and n_(2) are integers. Which of the following statements (s), is (are correct) 1. As wave length decreases the lines in the series converge 2. The integer n_(1) is equal to 2. 3. The ionisation energy of hydrogen can be calculated from the wave numbers of three lines. 4. The line of shortest wavelength corresponds to = 3.