Answer:
(a) 5.04 eV (B) 248.14 nm (c) 
Explanation:
We have given Wavelength of the light \lambda = 240 nm
According to plank's rule ,energy of light


Maximum KE of emitted electron i= 0.17 eV
Part( A) Using Einstien's equation
, here
is work function.
= 5.21 eV-0.17 eV = 5.04 eV
Part( B) We have to find cutoff wavelength



Part (C) In this part we have to find the cutoff frequency

Answer:multiplying will give us 7 significant figures and addition will give us 3 significant figures
Explanation:
After multiplying the two numbers they resulting value will give a value in its 4 decimal places because both given values are in 2 decimal places. The 4 dp is gotten by the addition of the decimal places of both given numbers (2+2) and
The result of its addition will give us a value in its 1dp and 3 significant figures since the addition of 23.68 and 4.12 will give us 27.8
They all stay the same regardless
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m