Question

Chapter 7 (Quantum Theory and Atomic Structure) 7 3 attempts left Check my work Be sure to answer all parts. Cobalt-60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this isotope has an energy of 5.70 MeV (million electron volts 1 eV- 1.602 x 10 J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray? Enter your answers in scientific notation. 10 points Frequency: eBook 10 Hz Print Wavelength: x 10 m References

Answer 1

Solution ☆ As you know that, where Ez energy h = planet's constant (6.626x10 347.5) 2= frequency Given that E = 5.70M eV And lev= 1.60281019 E = 5:70X108 x 1.602x109 = 9.13 x 1013 (:1 Meu=10 ev) Therefore -34 9.13810 J = 6.626x10 Jis Xo 0 = 9.13 x1013 6.626xlo 34 Jos = 1.38%16' 5! [=1138x10 Hz And we also know that E = Energy where, e. h = planck's constant c=speed of light I wave length -34 6.626x10 Js x 2.998x10 m/s 9.13×1513 2017 x 10-13 m :T =2.1781013 Ans)

solution © we know that Redberg equation T= Rp 1. 1 h where = wave length RH = Rydberg constant - (1.09677576x10 m) n,&n₂ z integers & n2>n Given that n2 =4, h,=2 Therefore, = 1.09679576x10m 2 -T 52 ) Į 2 0.2056 x 107 int tea x= 0.2056x107 ml = 40864 x 107 m . 486.4 X160 m (15 ma inm) I ~ 486 nm Aus & Solution :- Given that h₂= 9, h = 6 Therefore + 21.09677576x107 ml (1/2 - 1 - 621 5 20.016 998 167 mit 10 MEIA z 1016926x107 ml = 59.080x107m x=59080 x 10lom Az 59080 A Je