Question

PROBLEM #6. In BALMER lines in Hydrogen atom in Bohr model. An electron makes transition from n=4 to n2 A. Find the energy of the emitted radiation (photon) in this transition in ev. B. Find the wave length, frequency, and its De Broglie momentum C. Can you make a guess of the color of this photon? D. FOR THE ELECTRON IN n=2 CALCULATE ITS SPEED, RADIUS FROM THE NUCLUS, LINEAR MOMENTUM, ANGULAR MOMENTUM, KINETIC ENERGY, TOTAL ENERGY, AND De Broglie WAVE LENGTH

Answer 1

A) Energy of emitted photon,

E_photon = 13.6*(1/nf^2 - 1/ni^2) eV

= 13.6*(1/2^2 - 1/4^2) eV

= 2.55 eV

B) use, E_photon = h*c/lamda

lamda = h*c/E_photon

= 6.626*10^-34*3*10^8/(2.55*1.6*10^-19) (since 1 eV = 1.6*10^-19 J)

= 4.87*10^-7 m (or) 487 nm

frequency, f = c/lamda

= 3*10^8/(4.87*10^-7)

= 6.16*10^14 Hz

Debroglie momentum, p = h/lamda

= 6.626*10^-34/(4.87*10^-7)

= 1.36*10^-27 kg.m/s

c) Blue

d) v_n = 2.18*10^6/n m/s

if n = 2,

v2 = 2.18*10^6/2

= 1.09*10^6 m/s

radius of orbit,
rn = 0.0529*n^2 nm

uf n = 2,

r2 = 0.0529*2^2 nm

= 0.2116 nm (or) 2.116*10^-10 m

linear momentum, p = m*v

= 9.1*10^-31*1.09*10^6

= 9.92*10^-25 kg.m/s

angular momentum, L = r*P

= 2.116*10^-10*9.92*10^-25

= 2.10*10^-34 kg.m^2/s (or) J.s

kinetic energy = (1/2)*m*v^2

= (1/2)*9.1*10^-31*(1.09*10^6)^2

= 5.40*10^-19 J (or) 3.4 eV

Total energy = -13.6 eV/n^2

= -13.6/2^2

= -3.4 eV

debrglie wavelength, lamda = h/p

= 6.626*10^-34/(9.92*10^-25)

= 6.68*10^-10 m <<<<<<<<<<-----------Answer