**How do you find de Broglie wavelength with kinetic energy?** If you double the kinetic energy of a particle, how does the deBroglie wavelength change? Solution: λ = h/p, E = p^{2}/(2m), p is proportional to √E, l is proportional to 1/√E. **λ**_{2}/λ_{1} = √(E_{1}/E_{2}) = 1/√2.

**What is the de Broglie wavelength of an electron with kinetic energy?** Solution : Given, `KE = 120 eV, m = 9.1xx10^(-3)kg, e = 1.6xx10^(-19)c` <br> `lambda = (12.27)/(sqrt(V)) Å = (12.27)/(sqrt(120)) Å = 0.112xx10^(-9) m therefore lambda = **0.112 nm**`.

**How does kinetic energy relate to wavelength?** Definition: The Maximum Kinetic Energy of a Photoelectron given Wavelength. The maximum kinetic energy of a photoelectron is given by **???? = ℎ ???? ???? − ????** , m a x where ℎ is the Planck constant, ???? is the speed of light, ???? is the wavelength of the incident photon, and ???? is the work function of the metal surface.

**What is the de Broglie wavelength of an electron with a kinetic energy of 1.0 eV?** For an electron with KE = 1 eV and rest mass energy 0.511 MeV, the associated DeBroglie wavelength is **1.23 nm**, about a thousand times smaller than a 1 eV photon.

## How do you find de Broglie wavelength with kinetic energy? – Additional Questions

### What is the de Broglie wavelength of an electron with kinetic energy 50 EV?

Solution : `lambda=12.27/sqrt(v) =12.27/sqrt(50) =**1.732 Å**`.

### What is the de Broglie wavelength of an electron with kinetic energy 100ev?

=1.23 × 10−10 m = **1.** **23Ao**.

### What is the de Broglie wavelength of an electron that has 2.0 Kev of kinetic energy?

1 Answer. Truong-Son N. λ=**0.388 nm** .

### What is the wavelength of an electron that has 2.0 Kev of kinetic energy?

m = **0.0868 nm**.

### What will be the wavelength of an electron of energy 500 eV?

Solution : We are given : Energy of electron `= 500 eV= 500xx 1.6xx 10^(-19) J` <br> We know that, `lambda= (h)/(sqrt(2mE))= (6.62xx 10^(-34))/(sqrt(2xx 9.1xx 10^(-31)xx 500xx 1.6xx 10^(-19)))= 5.75xx 10^(-11)= **0.575 dot(A)`**.

### When the applied voltage is 10000 V the de Broglie wavelength of an electron is?

When the applied voltage is 10,000 V, the de-Broglie wavelength of an electron is. = **0.123 Å**.

### Is the kinetic energy of the particle is increased to 16 times its previous value the percentage change in the de Broglie wavelength of the particle is?

Solution : KE = 16 times `rArr` V = 4 times <br> `rArr` De-broglie wavelength `(lamda = (h)/(mV))` <br> = on fourth `(lamda_(0)rarr(lamda_(0))/(4))`. So De-Broglie wave will decrease by **75 %**.

### What is the wavelength of electron accelerated by a 10 volt?

λ=**1.** **23×10−9m**.

### What is the wavelength of a ball of mass 0.1 kg?

By using the above de Broglie equation, $lambda = dfrac{h}{{mv}} = dfrac{{6.626 times {{10}^{ – 34}}Js}}{{(0.1Kg) times 10m{s^{ – 1}}}} = 6.626 times {10^{ – 34}}m$ . Hence, **$6.626 times {10^{ – 34}}m$** is the answer.

### What will be the wavelength of a ball of mass 0.5 kg?

**1.32×10−35m**.

### What will be the wavelength of a ball of mass 100g?

λ=mvh=0. 1×1006. 626×10=**6.** **626×10−35m**.

### What will be the wavelength of a ball of mass 0.2 kg?

Answer: According to DE BROGLIE wavelength is inversely proportional to the momentum of the particle. =**2.652×10^-34 m** (aprox.)

### What is de Broglie wavelength of a ball of mass 0.12 kg?

The de-Broglie wave length of a ball of mass 0.12 kg is `**2.76xx10^-34m**.

### What is the de Broglie’s wavelength of a 50 kg mass moving at 20 m s?

1 Answer. Nam D. Truong-Son N. I get **2.65⋅10−36 m** .

### What is the de Broglie wavelength of a ball of mass 200 g?

λ=**2.** **385×10−30m**.

### What is the de Broglie wavelength of an electron with a kinetic energy of 12.8 eV?

Expert-verified answer

∴ K.E. = 12.8 × 10⁶ eV. ∴ v = 4.94 × 10⁷ m/s. ∴ λ = 7.99 × 10⁻¹⁵ m. ∴ λ = **7.99 fm**.

### What will be the wavelength of a ball of mass 0.2 kg moving with a velocity of 10ms 1?

Wave length =**6.62*10^-34m** Ans.