3. What is the energy of a photon whose frequency is 5.2 x 1015 Hz? Use the equation: E = hxv h= 6.626 10-34 J xs
4. What is the frequency of radiation whose wavelength is 2.40 x 105 cm? v=c+X
C= 3 x 108m/s

3. 3.45×10¯¹⁸ J.

4. 1.25×10¹⁵ Hz.

Explanation:

3. Determination of the energy of the photon.

Frequency (v) = 5.2×10¹⁵ Hz

Planck's constant (h) = 6.626×10¯³⁴ Js

Energy (E) =?

The energy of the photon can be obtained by using the following formula:

E = hv

E = 6.626×10¯³⁴ × 5.2×10¹⁵

E = 3.45×10¯¹⁸ J

Thus, the energy of the photon is 3.45×10¯¹⁸ J

4. Determination of the frequency of the radiation.

Wavelength (λ) = 2.4×10¯⁵ cm

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm

2.4×10¯⁵ cm = 2.4×10¯⁷ m

Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m

Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:

Wavelength (λ) = 2.4×10¯⁷ m

Velocity (c) = 3×10⁸ m/s

Frequency (v) =?

v = c / λ

v = 3×10⁸ / 2.4×10¯⁷

v = 1.25×10¹⁵ Hz

Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8