Question
The threshold wavelength for a photoelectric surface is
540 nm. Will a light of energy 2.5 eV cause photoemission?Solution
Let's calculate the value of ϕ in electron volts (eV). We know that hc ≈ 1240 eV nm. ϕ = 1240 eV nm / 540 nm = 2.3 eV The energy of the incident light is given as E=2.5eV. For photoemission to occur, the energy of the incident photon must be greater than or equal to the work function (E ≥ ϕ). Comparing the values: E=2.5eV ϕ≈2.3eV Since E>ϕ (2.5eV>2.3eV), photoemission will occur. \ Therefore, option (A) is correct.
√3598 × √(230 ) ÷ √102= ?
Simplify the following expressions and choose the correct option.
45% of 640 + (2/5 of 350) = ?
√225 + 27 × 10 + ? = 320
32 X 25 ÷ 4 + 12 of 30 = ? X 5 - 30Â
1440 ÷ 12 + 540 ÷ √36 + ? = 180 * 3
4567.89 - 567.89 - 678.89 = ?
- What will come in place of the question mark (?) in the following questions?
18×4+96÷8=? ((12+12+12+12)÷4)/((8+8+8+8+8+8)÷16) = ?
9999² + 1111² =?
- What will come in the place of question mark (?) in the given expression?
(120 - ?) ÷ 2 + 35 = 86 - 11