Answer:
E = 29.7× 10⁻²⁰ j
Explanation:
Given data;
Frequency of light = 4.48 × 10¹⁴ Hz
Energy of photon = ?
Solution:
Formula:
E = h.f
E = energy of photon
h = planck's constant
f = frequency
E = h.f
E = 6.63 × 10⁻³⁴ Kg.m² /s × 4.48 × 10¹⁴ s⁻¹
E = 29.7× 10⁻²⁰ Kg.m²/s²
Kg.m²/s² = j
E = 29.7× 10⁻²⁰ j
Answer:
b. Beta emission, beta emission
Explanation:
A factor to consider when deciding whether a particular nuclide will undergo this or that type of radioactive decay is to consider its neutron:proton ratio (N/P).
Now let us look at the N/P ratio of each atom;
For B-13, there are 8 neutrons and five protons N/P ratio = 8/5 = 1.6
For Au-188 there are 109 neutrons and 79 protons N/P ratio = 109/79=1.4
For B-13, the N/P ratio lies beyond the belt of stability hence it undergoes beta emission to decrease its N/P ratio.
For Au-188, its N/P ratio also lies above the belt of stability which is 1:1 hence it also undergoes beta emission in order to attain a lower N/P ratio.
Answer:
Explanation:
There are 3 types of plastids :-
1) Chloroplasts:- The green plastids which contain chlorophyll pigments for photosynthesis.
2) Chromoplasts:-The coloured plastids for pigment synthesis and storage.
3) Leucoplasts:- The colourless plastids for monoterpene synthesis found in non- photosynthetic parts of the plants.
They are of three types:-
a) Amyloplasts- stores starch.
b) Proteinoplasts- stores proteins.
c) Elaioplasts- stores fats and oils.
I would radiation a robotics arm but for a heated surphace is Shiny surfaces are poor absorbers and emitters (but they are good reflectors of infrared radiation).
Answer:
I think 767.8
Explanation:
I first did 10 times 22 bc that little carrot sign means squared so I did 10 times 22 and that equaled 220, then I did 3.49 times 220 and that is the answer which is 767.8
