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Energy of power,E = hv
(where h is plank's constant= 6.626176*10^-34 J)
v is frequency of radiation = 10^14 Hz
so, Energy,E = 6.62 * 10^-34 * 10^14
= 6.62 * 10^20
No.of proton = overall energy/energy of single proton
= 6.62 / 6.62 * 10^-20
= 10^20
therefore, the no. of protons of frequency 10^14hz in radiation of 6.62 J will be 10^20
(where h is plank's constant= 6.626176*10^-34 J)
v is frequency of radiation = 10^14 Hz
so, Energy,E = 6.62 * 10^-34 * 10^14
= 6.62 * 10^20
No.of proton = overall energy/energy of single proton
= 6.62 / 6.62 * 10^-20
= 10^20
therefore, the no. of protons of frequency 10^14hz in radiation of 6.62 J will be 10^20
mitse0102:
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Answer:
Energy of power,E = hv
(where h is plank's constant= 6.626176*10^-34 J)
v is frequency of radiation = 10^14 Hz
so, Energy,E = 6.62 * 10^-34 * 10^14
= 6.62 * 10^20
No.of proton = overall energy/energy of single proton
= 6.62 / 6.62 * 10^-20
= 10^20
therefore, the no. of protons of frequency 10^14hz in radiation of 6.62 J will be 10^20
Explanation:
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