Solution - Career Point

3. Hydrogen atom in its ground state is excited bymeans of a monochromatic radiation of wavelength970.6 Å. How many different wavelengths arepossible in the resulting emission spectrum ? Find thelongest wavelength amongst these.Sol. Energy the radiation quantum−34hc 6.6×10 × 3×10E = hv = = λ−10−19970.6×10 × 1.6×10= 12.75 eVEnergy of the excited sateE n = – 13.6 + 12.75 = – 0.85 eV13.6Now, we know that E n = –2nor n 2 13.6 −13.6= – = = 16E n − 0.85or n = 4The number of possible transition in going to theground state and hence the number of differentwavelengths in the spectrum will be six as shown inthe figure.n4321The longest wavelength corresponds to minimumenergy difference, i.e., for the transition 4 → 3.Now E 3 = –hcλmax13.623= E 4 – E 3= – 1.51 eV−346.6×10 × 3×10or λ max =−19(1.51−0.85) × 1.6×10= 18.75 × 10 –7 m = 18750 Å4. X-rays are produced in an X-ray tube by electronsaccelerated through a potential difference of 50.0 kV.An electron makes three collisions in the targetbefore coming to rest and loses half its kinetic energyin each of the first two collisions. Determine thewavelengths of the resulting photons. Neglect therecoil of the heavy target atoms.88Sol. Initial kinetic energy of the electron = 50.0 keVEnergy of the photon produced in the first collision,E 1 = 50.0 – 25.0 = 25.0 keVWavelength of this photon−34hc 6.6×10 × 3×10λ 1 = =−193E 1 1.6×10 × 12.5×10= 0.99 × 10 –10 m= 0.99 ÅKinetic energy of the electron after third collision = 0Energy of the photon produced in the third collision ,E 3 = 12.5 – 0 = 12.5 keVThis is same as E 2 . Therefore, wavelength of thisphoton, λ 3 = λ 2 = 0.99 Å5. In an experiment on two radioactive isotopes of anelements (which do not decay into each other), theirmass ratio at a given instant was found to be 3. Therapidly decaying isotopes has larger mass and anactivity of 1.0 µCi initially. The half lives of the twoisotopes are known to be 12 hours and 16 hours.What would be the activity of each isotope and theirmass ratio after two days ?Sol. We have, after two days, i.e., 48 hours,N 1 =40⎛1 ⎞N1⎜ ⎟⎝ 2 ⎠3=0⎛1 ⎞N 2 = N 2⎜⎟ =⎝ 2 ⎠Mass ratio =0N 1 /160N 2 /80N 1 =10N 2 N 20 0Now, A 1 = λ 1 N 1 = 1.0 µCiAfter two days,ButorN 8 3× 8 3. = = 16 162 20A 1 = λ 1 N 1 = λ 1 N 1 /16 =0A 2 = λ 2 N 2 = λ 2 N 2 /8λλ21=T 1 = 16T2λ 2 = 43λ1⎛ 3A 2 = ⎜⎝ 4λ 112 = 43⎞ ⎛ 1 0 ⎞ 1⎟ × ⎜ N 1 ⎟ ×⎠ ⎝ 3 ⎠ 81 0 1 0= λ1 N 1 = A 132 32= (1/32) µCi80A 1 /16 = (1/16)µCiXtraEdge for IIT-JEE 29 APRIL 2010