Answer:
E = 15×10⁻²⁹ J
Explanation:
Given data:
Frequency of photon = 2.2× 10⁷ Hz
Energy of photon = ?
Solution:
Formula:
E = h.f
h = 6.63×10⁻³⁴ Js
by putting values,
E = 6.63×10⁻³⁴ Js × 2.2× 10⁷ s⁻¹
E = 14.586 ×10⁻²⁹ J
E = 15×10⁻²⁹ J
The energy of photon is 15×10⁻²⁹ J.
The last electron on the outer most shell. or d oder of dere atomic no
A. If an objects velocity is decreasing, the object is said to be decelerating not accelerating.
B. If an objects velocity changes, it is either experiencing acceleration or deceleration
C. If an object is said to be decelerating, its velocity must be decreasing.
D. If an objects velocity remains constant, its acceleration is zero.
∴ B is correct
Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)