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The first ionization energy of a nitrogen atom is 2.32 aJ (attojoules). What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?

User Dayananda
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1 Answer

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Step-by-step explanation:

It is given that,

The ionization energy of a nitrogen atom,
E=2.32\ aJ

aJ is attpjoules.
1\ aJ=10^(-18)\ J


E=2.32* 10^(-18)\ J

The energy of an atom is given by :


E=h* f

f is the frequency of the photons


f=(E)/(h)


f=(2.32* 10^(-18))/(6.67* 10^(-34))


f=3.47* 10^(15)\ Hz

Also,
E=(hc)/(\lambda)


\lambda=(hc)/(E)


\lambda=(6.67* 10^(-34)* 3* 10^8)/(2.32* 10^(-18))


\lambda=8.62* 10^(-8)\ m

Hence, this is the required solution.

User Liecno
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