Given: I = 1500 W/m²; Wavelength = 5000?
Required equation ➔ E=hv=\(\frac {hc}{\lambda }\)
Speed of light ( c ) = 3 × 10⁸ m/s
Number of photons/s received = n = \(\frac {IA}{E} = \frac {(1500 \times 1) \times (5000 \times 10^{-10})}{6.63 \times 10^{-34} \times 3 \times 10^8} \)
n = 3.770 × 10²¹
Therefore, number of electrons received per second is 3.770 × 10²¹.

Yes, this is a true statement. Solar panels work on basic principle that when light strikes cathode, it causes emission of electrons, which in turn will produce an electric current. Therefore, photoelectric effect finds application in solar panels.

K_max = \(( \frac {hc}{\lambda })\) ? W₀
K_max = \( [ \frac {(6.6 \times 10^{-34} \times 3 \times 10^8 )}{(6.2 \times 10^{-6} \times 1.6 \times 10^{-19}) }] \) ? 0.1.
K_max = 0.2 – 0.1
K_max = 0.1 eV

Light travels as a wave due to its electromagnetic nature. This is furr facilitated by Maxwell?s equations which states that changing magnetic fields give rise to electric fields and changing electric fields give rise to magnetic fields.

Given: m = 150 g; v = 50 m/s
The required equation ➔ λ=\(\frac {h}{p} = \frac {h}{mv}\)
λ=\(\frac {6.6 \times 10^{-34}}{150 \times 10^{-3} \times 50}\)
λ=8.8 × 10^-35 m

Ultraviolet rays are most effective for photoelectric emission because y have highest frequency when compared to or electromagnetic waves, and hence most energetic among all of m.

Threshold frequency is given as:
v₀ = \(\frac {W_0}{h}\)
v₀ = \(\frac {W_0}{h}\)
v₀ = \(\frac {(3.3 \times 10^{-19})}{(6.6 \times 10^{-34})}\)
v₀ = 5 × 10¹⁴ Hz.

Yes, this is a true statement. Thermionic emission is process of electron emission involving liberation of electrons from an electrode by virtue of its temperature and this is done by releasing energy supplied by heat. This happens as rmal energy given to charge carrier overcomes work function of material.

Davisson ? Germer experiment uses an electron gun to produce a fine beam of electrons which can be accelerated to any desired velocity by applying a suitable voltage across gun. The ors mentioned do not find an application here.

According to equation, W₀ = hv₀ = \(\frac {hc}{\lambda_0}\), work function is inversely proportional to wavelength.
So metal A with lower work function has a higher threshold wavelength.