Answer:
E = 3.6×10⁻¹⁹ J
Explanation:
Given data:
Wavelength = 550 nm (550 ×10⁻⁹ nm)
Energy of wave = ?
Solution:
Formula:
E = h c/λ
c = 3×10⁸ m/s
h = 6.63×10⁻³⁴ Js
Now we will put the values in formula.
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s /550 ×10⁻⁹ nm
E = 19.89×10⁻²⁶ J.m /550 ×10⁻⁹ nm
E = 0.036×10⁻¹⁷ J
E = 3.6×10⁻¹⁹ J
<span>Photons were the first sub-atomic particles detected, but not quite discovered as they could not be explained. Photons were first detected by Johann Wilhelm Ritter, Victor Schumann, and Winhelm Rontgen. The next, and first sub-atomic particle discovered, was the electron. The electron was discovered by J. J. Thompson in the late 1800s. The next two sub-atomic particle discoveries were the alpha particle and photon, discovered by Ernest Rutherford and Paul Villard respectively. Rutherford also discovered the proton and in 1932, James Chadwick discovered the neutron.</span>
Answer: (1,3-dimethylbutyl) acetate
Explanation:
sorry if it wrong
To create the liquid and superfluid states you cool down helium gas to a few degrees above absolute zero
The energy released when electron move from n=4 to n=3 is 0.66 eV
We know that in an atom energy of nth state is
eV
where n is the energy level
Therefore,
Thus, 
= -0.85eV
= -1.51eV
Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -
= -0.85 - ( -1.51)
= 0.66eV
To know more about energy released in electron transition
brainly.com/question/8384785
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