An electron in $$hydrogen-like$$ atom makes a transition from nth orbit and emits radiation corresponding to Lyman series. If $$de-Broglie$$ wavelength of electron in nth orbit is equal to the wavelength of radiation emitted, find the value of n. The atomic number of atom is $$11$$.

Question

An electron in $$hydrogen-like$$ atom makes a transition from nth orbit and emits radiation  corresponding to Lyman series. If $$de-Broglie$$ wavelength of electron in nth orbit is equal to the wavelength of radiation emitted, find the value of n. The atomic number of atom is $$11$$.

Infinity Learn · Accepted Answer

If A is $$de-Broglie$$ wavelength, then for nth stationary $$orbit2$$$$π$$$$rn=n$$λwhere rn is radius of nth orbit.$$rn=$$ε$$0h2n2$$$$π$$$$mZe2$$∴ $$2$$πε$$0h2n2$$$$π$$$$mZe2=n$$λ⇒$$1$$λ$$=mZe22$$ε$$0h2n$$    ........(i)For$$ Lyman series of hydrogen like atom, wavelength is given as,$$1$$λ$$=Z2R112$$−$$1n2$$    ........$$(ii)From$$ Eqs. $$) and (i), $$Z2R1$$−$$1n2=mZe22$$ε$$0h2nRydberg$$ constant, $$R=me48$$ε$$02ch3$$∴    $$Z2me48$$ε$$02ch31$$−$$1n2=mZe22$$ε$$0h2n$$ or     $$1$$ $$1$$−$$1n2=4$$ε$$0chm2ZSubstituting$$ given values, we $$get=4$$×$$8.85$$×$$10$$−$$12$$×$$3$$×$$108$$×$$6.62$$×$$10$$−$$34n$$×$$1.6$$×$$10$$−$$192$$×$$11=25n$$∴ $$n2$$−$$1=25n$$ or  $$n2$$−$$25n$$−$$1=0n$$−$$25$$±$$($$−$$25)2+4$$×$$1$$×$$12$$≈$$25$$±$$6252$$≈$$25$$ As negative n is not possible, n≈$$25Alternative$$ solution  :     Energy of emitted photon $$=$$ $$=$$ hCλ$$=$$ $$13.6$$ $$z2(1-1n2)$$   where λ $$=$$ wavelength of emitted photon                $$=de$$ broglie wavelength in n th orbit $$=$$ $$3.3nzA0Solving$$ n $$=$$ $$25$$

An electron in hydrogen-like atom makes a transition from nth orbit and emits radiation corresponding to Lyman series. If de-Broglie wavelength of electron in nth orbit is equal to the wavelength of radiation emitted, find the value of n. The atomic number of atom is 11.

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