What is the approximate frequency of a photon having an energy 3.2 * 10^-24 j? (h = 6.6*10^-34 j*s) a.) 48*10^9 b.) 2.1*10^-57 c.) 3.0*10^-58 d.) 2.1*10^-10

The Planck’s Equation states that

E = hv

E stands for energy measured in Joules

v stands for frequency measured in s^-1 or Hertz (Hz)- 1Hz .

h is Planck’s constant which is equivalent to 6.626x10-34 J.s

The Planck's equation states that the energy of a photon, is directly proportional to its frequency (v), by a constant factor (h).

Given:

E=3.2 * 10^-24 J

h= 6.626x10-34 J.s

Now we know that

E=hv

Thus substituting the given values we get

v=3.2 * 10^-24 /6.626x10-34

v=48 x 10^9 s^-1

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.  

for example, fusion of hydrogen atoms at the sun.  

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).  

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.  

the objects have the same charge is correct.

nathan lee