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umka2103 [35]
3 years ago
5

When an atom is involved in a nuclear reaction:

Physics
1 answer:
nataly862011 [7]3 years ago
4 0
I believe the answer would be , it may change from one element to another. I’m saying this because during nuclear transmutation is when a subatomic particle fired at the nucleus of an atom changes into a heavier element , to break the nucleus apart into two nuclei and energy.
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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

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