For the Paschen series thr values of n(1...

For the Paschen series thr values of `n_(1)` and `n_(2)` in the expression `Delta E = R_(H)c [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` are

`n_(1) = 1, n_(2) = 2, 3, 4`

`n_(1) = 2, n_(2) = 3, 4, 5`

`n_(1) = 3, n_(2) = 4, 5, 6`

`n_(1) = 4, n_(2) = 5,6,7`

Text Solution

Verified by Experts

The correct Answer is:

Topper's Solved these Questions

ATOMIC STRUCTURE
DINESH PUBLICATION|Exercise Selected Straight Objective Type MCQs|70 Videos
ATOMIC STRUCTURE
DINESH PUBLICATION|Exercise Compre.|17 Videos
ATOMIC STRUCTURE
DINESH PUBLICATION|Exercise Ultimate Preparatory Package|15 Videos
APPENDIX
DINESH PUBLICATION|Exercise Completion Reactions|15 Videos
BIOMOLECULES
DINESH PUBLICATION|Exercise Reason|1 Videos

Similar Questions

Explore conceptually related problems

For a single electron atom or ion the wave number of radiation emitted during the transition of electron from a higher energy state (n = n_(2)) to a lower energy state (n=n_(1)) is given by the expression: bar(v) = (1)/(lambda) = R_(H).z^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) ...(1) where R_(H) = 2(pi^(2)mk^(2)e^(4))/(h^(3)c) = Rydberg constant for H-atom Where the terms have their usual meanings. Considering the nuclear motion, the most accurate expression would have been to replace mass of electron (m) by the reduced mass (mu) in the above expression, defined as mu = (m'.m)/(m'+m) where m'= mass of nucleus For Lyman series: n_(1) =1 (fixed for all the lines) while n_(2) = 2,3,4 ... for successive lines i.e. 1^(st), 2^(nd),3^(rd) ... lines, respectively. For Balmer series: n_(1) = 2 (fixed for all the lines) while n_(2) = 3,4,5 ... for successive lines. The ratio of the wave numbers for the highest energy transition of e^(-) in Lyman and Balmer series of H-atom is:

For a single electron atom or ion the wave number of radiation emitted during the transition of electron from a higher energy state (n = n_(2)) to a lower energy state (n=n_(1)) is given by the expression: bar(v) = (1)/(lambda) = R_(H).z^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) ...(1) where R_(H) = 2(pi^(2)mk^(2)e^(4))/(h^(3)c) = Rydberg constant for H-atom Where the terms have their usual meanings. Considering the nuclear motion, the most accurate expression would have been to replace mass of electron (m) by the reduced mass (mu) in the above expression, defined as mu = (m'.m)/(m'+m) where m'= mass of nucleus For Lyman series: n_(1) =1 (fixed for all the lines) while n_(2) = 2,3,4 ... for successive lines i.e. 1^(st), 2^(nd),3^(rd) ... lines, respectively. For Balmer series: n_(1) = 2 (fixed for all the lines) while n_(2) = 3,4,5 ... for successive lines. If proton in H-nucleus be replaced by positron having the same mass as that of electron but same charge as that of proton, then considering the nuclear motion, the wavenumber of the lowest energy transition of He^(+) ion in Lyman series will be equal to

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

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?

The emission spectrum of hydrogen is found to satisfy the expression for the energy change Delta E (in joules) such that Delta E = 2.18 xx 10^(-18)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))J where n_(1) = 1,2,3,.... and n_(2) = 2,3,4,... The spectral lines correspond to Paschen series if

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. 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:

DINESH PUBLICATION-ATOMIC STRUCTURE-Revision Questions

Which of the following is not possible for 4p or 3d electrons?
Text Solution
|
Which of the following represents the molarity of pure water?
Text Solution
|
For the Paschen series thr values of n(1) and n(2) in the expression D...
Text Solution
|
What is the atomic number of the element with M^(2+) ion having electr...
Text Solution
|
The correct set of four quantum numbers for outermost electrons of pot...
Text Solution
|
A body of mass x kg is moving with a velocity of 100 ms^(-1). Its de-B...
Text Solution
|
Number of unpaired electrons in Fe^(3+) (Z = 26) is
Text Solution
|
The correct set of quantum number for the unpaired electron of a chlor...
Text Solution
|
Common salt obtained from sea water contains 95% NaCl by mass. The app...
Text Solution
|
Whith increasing principal quantum number, the energy difference betwe...
Text Solution
|
Wavelength of electron waves in two Bohr orbits is in ratio3:5 the rat...
Text Solution
|
The energy of an electron in the 3rd orbit of an atom is -E. The energ...
Text Solution
|
The e//m ratio for electron was determined by ……..
Text Solution
|
Which orbital give an electron the greatest probability of being found...
Text Solution
|
If the de Broglie wave length of the fourth Bohr's orbit of hydrogen a...
Text Solution
|
From the given sets of quantum numbers, the one that is inconsistent w...
Text Solution
|
A particle having a mass of 10 milligram has a velocity of 3600 km/hou...
Text Solution
|
For the principal quantum number n = 2, the possible values of azimuth...
Text Solution
|
Which is the correct order of wave number of the following radiations:...
Text Solution
|
What is the maximum number of emission lines when the excited electro...
Text Solution
|