Home
Class 12
PHYSICS
A: Spectral analysis can differentiate b...

A: Spectral analysis can differentiate between isotopes as per the equation `1/lambda = RZ^(2)[1/(n_(1)^(2))- 1/(n_(2)^(2))]`.
R: Rydberg's constant R is not a universal constant and is dependent on the mass of nuclei as well.

A

If both Assertion & Reason are true and the reason is the correct explantion of the assertion , then mark (1)

B

if both Assertion & Reason are true but the reason is not the correct explantion of the assertion , then mark (2)

C

If Assertion is true statement but Reason is false, then mark (3)

D

If both Assertion and Reason are false statements, then mark (4)

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Similar Questions

Explore conceptually related problems

The emission series of hydrogen atom is given by (1)/(lambda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) where R is the Rydberg constant. For a transition from n_(2) to n_(1) , the relative change Deltalambda//lambda in the emission wavelength if hydrogen is replaced by deuterium (assume that the mass of proton and neutron are the same and approximately 2000 times larger than that of electrons) 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 What will be the value of modified Rydberg's constant, if the nucleus having mass m_(N) and the electron having mass m_(e) revolve around the centre of the mass?

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 energy of n^(th) orbit is given by E_(n) = ( -Rhc)/(n^(2)) When electron jumpsfrom one orbit to another orbit then wavelength associated with the radiation is given by (1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2))) The series that belongs to visible region is

Balmer gives an equation for wavelength of visible radition of H^(-) spectrum as lambda=(kn^(2))/(n^(2)-4) .The value of k in terms of Rydberg's constant R is

Balmer gave an equation for wavelength of visible radiation of H-spectrum as lambda=(kn^(2))/(n^(2)-4) . The value of k in terms of Rydberg's constant R is m//R , where m is :

sum_(r=1)^(n)(r)/(2^(r))=2-((1)/(2))^(2)(n+2)

The wavelength of the various lines in the hydrogen atomic emission spectrum can be related by a mathematical equation: (1)/(lambda) = R ((1)/(n^(2))-(1)/(m^(2))) What is the value of n for the Pfund series of lines?

Differentiate w.r.t.x(2x+1)^(3)

Differentiate the following w.r.t. x: |2x - 1|